Find the mi of a uniform semicircular disc of mass m and radius r about an axis perpendicular If `T` is the time period of small osciallations about an asked Jun 26, 2019 in Physics by Anshuman Sharma ( 78. A uniform semicircular disc of mass \'m\' and radius \'R\' is shown in the figure. Radius of each disk = R. Find out the moment of inertia of a semicircular disc about an axis passing through its centre of mass and perpendicular to the plane? (mass=M and radius=R). Taking point A as the origin, the x-axis along AB and the y-axis perpendicular to AB, the centre of mass of the disc is located at co-ordinates (r,4π/3π). of a uniform semicircular disc of mass M and radius 'R' about an axis perpendicular to its plane and passing through point 'P' as shown. Its moment of inertia about an axis parallel to this axis but passing through the edge of the disc is (see the given figure. Calculate the radius of gyration of a uniform circular disk of radius r and thickness t about a lim perpendicular to the plane of this disk and tangent to the dinle A uniform semi-circular disc of mass M and radius R lies in the X − Y plane with its centre at the origin as shown in the figure. Find out its moment of inertia about. about an axis passing through a point on its circumference and perpendicular to its plane is (mass of circular disc = 5 k g, radius of disc = 1 m) approximately: The moment of inertia of a uniform semicircular disc of mass `M` and radius `r` about a line perpendicular to the plane of the disc through the center is Mass of a disc is 2 kg and radius of gyration of the disc about a transverse axis passing through its centre of mass is 5 0 cm. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is : (1) 40/9 MR 2 (2) 10 MR 2 (3) 37/9 MR 2 (4) 4 MR 2 May 31, 2019 · A uniform semi-circular disc of mass `m` and radius `r` is suspended as shown in the figure. 1/4 Mr 2D. A. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is. r. lang. An unstretchable rope is found on the disc. `(MR^(2))/(2) if theta =0^(@)` C. A uniform semicircular disc of mass `\'m\'` and radius `\'R\'` is shown in the figure. (a) 2 (b) 3 (c) 4 (d) 5 (b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be M R 2 / 4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge. (c) A uniform square plate of edge a suspended through a corner. A chip of mass m breaks off the edge of the disk at an instant such that A circular portion of diameter R is cut out from a uniform circular disc of mass M and radius R as shown in figure. Two uniform semicircular discs, each of radius R, are stuck together to form a disc. I of a semi circular ring of mass M and radius ′ R ′ about an axis lying the plane which is passing through its centre and angle θ with the line joining its end is: M R 2 2 if θ = 0 o; M R 2 2 at θ = 60 o; M R 2 2 at θ = 120 o; M R 2 2 at θ = 180 o Click here👆to get an answer to your question ️ Find the M. Moment of Inertia: Disc and Washer (a) A thin uniform disc of mass M and radius Ris mounted on an axis passing through the center of the disk, perpendicular to the plane of the disc. Three forces act in the +y-direction on the disk: 1) a force 316 N at the edge of the disk ; A uniform disk with mass m = 9. Masses of the two semicircular parts are M and 3 M. 16. (Given that moment of inertia of removed portion about the given axis is M R 2 / 2) The M. 181 2 M R 2. 6k points) Q. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is From a semi-circular disc of mass M and radius R 2. View Solution The moment of inertia of a uniform semi circular disc about the axis passing through the centre of mass and perpendicular to its plane is (Mass of the disc is M and radius is R) $\begin{align} & a)\dfrac{M{{R}^{2}}}{2}+M{{\left( \dfrac{4R}{3\pi } \right)}^{2}} \\ & b)\dfrac{M{{R}^{2}}}{2}-M{{\left( \dfrac{4R}{3\pi } \right)}^{2}} \\ The moment of inertia of a semicircular ring about a line perpendicular to the plane of the ring through its centre is given as $I=m{{r}^{2}}$, where m and r are the mass and radius of the ring. 4/042-1 G nut ac DRE. From a semi-circular disc of mass M and radius${{R}_{2}}$. The moment of inertia of a uniform semi circular disc about an axis passing through its centre of mass and perpendicular to its plane is (Mass of this disc is M and radius is R) View Solution Q 4 From a semi-circular disc of mass M and radius R 2. Also, axis P Q is in the X Y plane. Its moment of inertia about an axis through its centre of mass and perpendicular to its plane is The correct answer is We know that the moment of inertia of a circular disc about an axis perpendicular to the plane of disc and passing through centre is (1/2) Mr2. 31 m lies in the x-y plane and centered at the origin. Find the M. The moment o inertia of a circular disc of radius 2 m and mass 1 k g about an axis passing through the center of mass but perpendicular to the plane of the disc is 2 k g − m 2. m. A uniform cylinder has a radius R and length L. The moment of inertia of a uniform semi circular wire of mass m and radius r, about an axis passing through its centre of mass and perpendicular to its plane is mr? (1--2). Ans: Hint: Find the moment of inertia of the semicircular disc about the axis passing throug A disc of radius R, is cut from a disc of radius R 2 as shown in figure. Mass of this disc is M and radius is R. RELATED QUESTIONS. What is its kinetic energy? Its angular momentum? B. The center of the ring is at a distance of d = √ 3 R from the center of the sphere. 40/4 M R 2C. Suggest Corrections 28 Two identical uniform disc of mass m and radius r are arranged as shown in the figure. Find the moment of inertia of a uniform half-disc about an axis perpendicular to the plane and passin through its centre of mass. Find the moment of inertia of the disc about an axis passing through its centre and perpendicular to its plane. The radius of the disc is 1 decimetre and mass is A thin uniform disc of mass M and radius R has concentric hole of radius r. A semicircular ring of mass m and a semicircular disc of mass 2 m of same radius are joined together as shown. Click here👆to get an answer to your question ️ 39. -. The gravitational attraction between the sphere and ring is Thus, the moment of inertia of a uniform circular disc of radius R and mass M about an axis passing from the edge of the disc and normal to the disc is 3 2 M R 2. about a ciameter and about its tangent in a plane of disc. The moment of inertia of a uniform semicircular wire of mass m and radius r. `(a)` axis `\'AB\' (` shown in t 3 days ago · A uniform semicircular disc of mass `'m'` and radius `'R'` is shown in the figure. Determine the moment of Inertia of a uniform disc having mass M and radius R about an axis passing through centre & perpendicular to the plane, of the disc? The kinetic energy of a circular disc rotating with a speed of 60 r. So, moment of inertia of the disc with A thin disc of mass M and radius R has mass per unit area σ (r) = k r 2, where r is the distance from its centre. 1. The moment of inertia of a circular ring of mass M, radius R about an axis perpendicular to its plane and passing through its centre is: Q. Determine the moment of Inertia of a uniform disc having mass M and radius R about an axis passing through centre & perpendicular to the plane, of the disc? From a uniform circular disc of radius R and mass 9 M, a small disc of radius R 3 is removed as shown in the figure. 2/5MR^2 C. 5 m. about a line, in its plane, passing through its centre and inclined at an angle o to the major axis is -m (b? cos? e + a² sin’e) (B) The M. 1/2 Mr 2 The moment of inertia of a uniform semi circular disc about an axis passing through its centre of mass and perpendicular to its plane is (Mass of this disc is M and radius is R) Q. of a uniform semicircular disc of mass M and radius R about a line perpendicular to the plane of the disc and passing through the centre is a)-MR2 b) MR c) MR2 d) MR2 241. MR^2/2 The moment of inertia of a uniform semi circular disc about an axis passing through its centre of mass and perpendicular to its plane is (Mass of this disc is M and radius is R) View Solution Q 3 Apr 29, 2024 · The moment of inertia of a uniform semicircular ring with mass m and radius r about an axis perpendicular to the plane and through the center is related to, but directly given by, the known moment of inertia of a full ring (mr^2) and a semicircular lamina (ma^2 / 4). Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is A thin disc of mass M and radius R has mass per unit area σ (r) = k r 2, where r is the distance from its centre. = 2/5 MR'] 17. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is: From a semi-circular disc of mass M and radius R 2. more. A disc of radius R 3 is cut as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is : Click here👆to get an answer to your question ️ Compound pendulums are made of: (a) A tod of length & suspeonded through a point located at distance (14 from centre of rod. i) = W ii) lo = MR2 aalid aulindar of uniform density has a length equal to ite diam atau . Find the moment of inertia of the remaining disc about an axis passing through 0 and perpendicular to the plane of disc. May 31, 2019 · The moment of inertia of a uniform semicircular disc of mass `M` and radius `r` about a line perpendicular to the plane of the disc through the center asked Jun 7, 2019 in Physics by Navinsingh ( 86. From a uniform circular disc of radius R and mass 9 M, a small disc of radius R 3 is removed as shown in the figure. 4 MR 2B. For semi-circular discI=1212(2M)r2=12Mr2 The moment of inertia of a uniform semi-circular disc of mass m and radius r about a line perpendicular to the plane of the disc passing through center is The moment of inertia of a uniform semi-circular disc of mass m and radius r about a line perpendicular to the plane of the disc passing through center is From a circular disc of radius R and mass 9 M, a small disc of radius R / 3 is removed from the disc. The moment of inertia of a uniform semicircular disc of mass M and radius R about a line perpendicular to the plane of the disc and passing through the center is: A. calculate its total energy. ) The moment of inertia of a thin, uniform, flat semicircular disc of mass m and radius r for rotation about its diameter AB is mr2/4. (d) A uniform disc of mass m and radius r suspended through a point r/2 A thin uniform circular disc of mass M and radius R is rotating with an angular velocity (ω) in a horizontal plane about a vertical axis passing through its centre and perpendicular to its plane. 37/9 M R 2 A ring and a solid sphere of same mass and radius are rotating with the same angular velocity about their diametric axes then: Then (-a, 0) 0 (a,0)^ (0, -b) (A) The M. A racing car comnetes 5 rounds of a circular track in 2 minute. Step 1: Given Data: Number of disks = 7. In this case the mass of the disc is 2M. A uniform semicircular disc of mass `'m'` and radius `'R'` is shown in the figure. If the moment of inertia of this cylinder about an axis passing through its centre and normal to its circular face is equal to the moment of the same cylinder about an axis passing through its centre and perpendicular to its length, then A thin disc of mass M and radius R has mass per unit area σ (r) = k r 2, where r is the distance from its centre. The angular velocity of the system is: A semicircular 2-dimensional plate of mass m has radius r and centre C. I. Feb 17, 2020 · Two uniform semicircular discs, each of radius `R`, are stuck together to form a disc. if the disc left has mass M then find the moment of inertia of the disc left about an axis passing through its centre and perpendicular to its plane Click here👆to get an answer to your question ️ Q. If string does not slip on the rim then find the acceleration of the ning (or) Calculate the moment of inertia of uniform circular disc of mass 500g, radius 10 cm about, (i) the diameter of the disc (a) the axis, tangent to the disc and parallel to its diameter and (iii) the axis through the centre of the disc and perpendicular to its plane. The angle through which the disk has turned varies with time according to theta(t) = (1. The block is released from rest. Hence, MOI of disc about the axis of rotation I = I c + m (r / 2) 2 [using parallel axis theorem] I = m r 2 2 + m r 2 4 and distance of COM of disc from the axis of rotation, d = r 2 ⇒ T = 2 π ⎷ 1 2 m r 2 Apr 6, 2019 · One quarter sector is cut from a uniform circular disc of radius R. s. View Solution Q 2 Sep 21, 2021 · Find the M. If α is the angular acceleration of the lower disc and a c m is acceleration of centre of mass of the lower disc, then relation among a c m , α , r is: The moment of inertia of a uniform semi circular disc about an axis passing through its centre of mass and perpendicular to its plane is (Mass of this disc is M and radius is R) Q. A semi-circular disc of radius R 1 is removed as shown in the figure. Find the value of k. `(MR^(2))/(2) if theta=90^(@)` Find the M. Jun 22, 2019 · Two uniform semicircular discs, each of radius `R`, are stuck together to form a disc. Hence, the moment of inertia of the complete disc is $\dfrac{(2M){{r}^{2}}}{2}=M{{r}^{2}}$. (b) A ring of mass m and radius r suspenided through a point on its periphery. 1500 kg m 2C. The moment of inertia of a circular disc about an axis perpendicular to its plane and passing throug its centre is equal to $\dfrac{M{{r}^{2}}}{2}$, where M is the mass of the disc and r is the radius of the disc. Find the momentum of inertia of the residual disc about an axis passing through centre O and perpendicular to the plane of the disc. Taking point A as the origin, the x-axis along AB and the y-axis perpendicular to AB, the centre of mass of the disc is located at co-ordinates (r, 4r/3p). = 1/2 mass x (radius)?] 16. The moment of inertia of a thin, uniform, flat semicircular disc of mass m and radius r for rotation about its diameter AB is mr. This sector has mass M. The induced emf between the inner and the outer edge of the annular disk is : A thin uniform disc of mass 9 M and of radius R. 09 kg and radius R = 1. A solid of mass 10 kg rolls without slipping with a velocity of 50 cm/sec. MR^2 D. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is The moment of inertia of a uniform semicircular disc of mass M and radius R about a line perpendicular to the plane of disc and passing through the centre is : A uniform ring of mass M and radius R is placed directly above a uniform sphere of mass 8 M and of radius R. The moment of inertia of a uniform semi circular disc about an axis passing through its centre of mass and perpendicular to its plane is (Mass of this disc is M and radius is R) Q. According to parallel axis theorem-I o = I c m + M d 2. (a) axis \'AB\' ( shown in the figure The correct option is D 3 2 R Here, moment of inertia of disc about a point on rim and perpendicular to plane of disc is I 0 = 1 2 M R 2 + M R 2 Time period of a physical pendulum. Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR 2 /5, where M is the mass of the sphere and R is the radius of the sphere. about a chord of length 2r passing through O and lying in plane of elliptic disc is 1 Ma²b2 4 p2 (C) The M. From a semi-circular disc of mass M and radius R 2. In this case, the mass of the half-ring is dm and its radius is x. 1000 kg m 2 The moment of inertia of a uniform semi circular disc about an axis passing through its centre of mass and perpendicular to its plane is (Mass of this disc is M and radius is R) Q. The approach for solving this question should be to find the moment of inertia of a circular disc first, and then the moment of inertia of a semicircular disc is found. if the disc left has mass M then find the moment of inertia of the disc left about an axis passing through its centre and perpendicular to its plane From a uniform circular disc of radius R and mass 9 M, a small disc of radius R 3 is removed as shown in the figure. of the section shown (a part of circular disc) about an axis perpendicular to its plane and passing through point 'O'. (Given that moment of inertia of removed portion about the given axis is M R 2 / 2) An annular circular brass disk of inner radius r and outer radius R is rotating about an axis passing through its center and perpendicular to its plane with a uniform angular velocity ω in a uniform magnetic field of induction B normal to the plane of the disk. MR^2/4 B. Find the radius of gyration of a uniform disc of radius R and mass M about its edge and perpendicular to the disc. Let the semicircular disc of mass M and radius R have a uniform mass density σ. A disc of radius R, is cut from a disc of radius R 2 as shown in figure. of a uniforin disc about and axis passing through its centre of mass and perpendicular to its plane is Hence find out the M. Nov 6, 2021 · The moment of inertia of a semicircular ring of mass M and radius R about an axis which is passing through its centre and at an angle `theta` with the line joining its ends as shown in figure is A. A disc of mass m and radius R is free to rotate in a horizontal plane about a vertical smooth fixed axis passing through its centre. The moment of inertia of the remaining (shaded) portion of the disc about an axis passing through the centre O of the disc and perpendicular to its plane is: A uniform disc of radius R and mass M is free to rotate about a fixed horizontal axis perpendicular to its plane and passing through its centre. \( \dfrac{ma^{2}}{4} \), since the mass distribution with respect to rotation about the diameter is the same. If the mass of original uncut disc is M, find the momentum of inertia of residual disc about an axis passing through centre O and perpendicular to the plane of the disc. Find the MOI of a uniform circular disc of mass M =10 kg and radius R =10 m about tangent perpendicular to its plane. Find the MOI about an axis passing through point O and perpendicular to the plane. Tangential acceleration of a point P on the periphery of the disc when a uniform force F is applied on the rope is Mar 18, 2022 · You are asking at Today. Determine the moment of inertia about the following two parallel axes that are perpendicular to the plane of the disk. about a line perpendicular to the plane of disc through the centre is: View Solution A semicircular disc of mass M and radius R is free to rotate about its diameter. Then, radius of the disc will be: Mar 29, 2018 · The moment of inertia of a uniform semicircular disc of mass M and radius r about a line perpendicular to the plane 2) 2/5Mr2 (3) Mr2 (4) 1/2Mr2 B 12 B-3. Mr 2C. Question Bank with Solutions Maharashtra State Board Question Bank with Solutions (Official) issing Q. Find the moment of inertia for a semicircular disc of mass m and radius r about an axis passing through its centre of mass and perpendicular to the plane of disc Q. `(MR^(2))/(2)if theta =` any angle D. 10 r; A uniform flat disk of mass M and radius R rotates about the horizontal axis through its center with angular speed \omega_o. 7k points) Hint: Moment of inertia is defined as the product of the mass of section and the square of the distance between the reference axis and the centroid of the section. Find the moment of inertia of the circular disc about an axis perpendicular to its plane and passing through its centre of mass. of a half disc of mass M and radius R about its 279 + diameter LOU (A) = MR10) MR” (C) MR? (D) XMR 4 p space for rough work Aug 7, 2019 · The moment of inertia of a uniform semicircular wire of mass M and radius r about a line perpendicular to the plane of the wire through the centre is asked Mar 24, 2018 in Physics by shabnam praween ( 138k points) Two uniform semicircular discs, each of radius R, are stuck together to form a disc. A thin uniform disc of mass M and radius R has concentric hole of radius r . Oct 3, 2023 · Find the M. A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Its moment of inertia about the axis of rotation is… (A) (1/2)MR 2 (B) (1/4)MR 2 (C) (1/8)MR 2 (D) √2MR 2 Click here👆to get an answer to your question ️ Moment of inertia of a uniform quarter disc of radius R and mass M about an axis through its centre of mass and perpendicular to its plane is: (A) MRGR The moment of inertia of a uniform semicircular disc of mass M and radius r about a line perpendicular to the plane of the disc through the centre is Q. 36 m lies in the x-y plane and centered at the origin. 15 A uniform disc of mass 5 kg has a radius of 0. A string is wrapped over its rim and a block of mass m is attached to the free end of the string. Jun 7, 2019 · Moment of inertia of a thin semicircular disc `(mass - M & radius = R)` about an axis through point `O` and perpendicular to plane of disc, is given b asked May 31, 2019 in Physics by Navinsingh ( 86. if the disc left has mass M then find the moment of inertia of the disc left about an axis passing through its centre and perpendicular to its plane Click here👆to get an answer to your question ️ (D) H 72 14. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is The moment of inertia of a uniform semicircular lamina of mass \( m \) and radius \( a \) about its base, or diameter, is also \( \dfrac{ma^{2}}{4} \), since the mass distribution with respect to rotation about the diameter is the same. Since it is given that the disc oscillates in a plane parallel to its plane, the axis of rotation must be perpendicular to its plane. moment of inertia of a uniform disc of mass M and radius R about an axis passing through its edge and perpendicular to its plane is I. p. B-4. 1400 kg m 2D. 1200 kg m 2B. ) From a uniform circular disc of radius R and mass 9 M, a small disc of radius R 3 is removed as shown in the figure. 10 M R 2D. A uniform disc mass m and radius r rotates along an axis passing through its centre of mass and normal to its plane. A uniform disc of mass m and radius R is pivoted at its centre O with its plane vertical as shown in figure, A circular portion of disc of radius R 2 is removed from it. about a line perpendicular to the plane of The moment of inertia of a uniform semi circular disc of mass M and radius . $\text{Mass density = }\dfrac{\text{Total mass of body}}{\text{cross sectional area}}$ Now, we know that the cross sectional area of a semicircular disk is $\dfrac{\pi {{R}^{2}}}{2}$ where R is the radius. T = 2 π √ I 0 M g d T = 2 π ⎷ 1 2 M R 2 + M R 2 M g R T = 2 π √ 3 R 2 g → (1) Also, Time period of simple pendulum, T = 2 π √ l g → (2) equating (1 The moment of inertia of uniform semicircular disc of mass M and radius r about a line perpendicular to the plane of the disc through the centre is From a uniform circular disc of radius R and mass 9 M, a small disc of radius R/3 is removed as shown in the figure. Then, Match for the moment of inertia of semi-circular disc about different axis. Calculate its moment of inertia about an axis passing through a point on its circumference and perpendicular to its plane. What is the radius of the track, if the car has Find the work done to make it rotate with double the speed. Find the moment Apr 7, 2019 · The moment of inertia of uniform semi-circular disc of mass M and radius r about a line perpendicular asked Nov 2, 2018 in Rotational motion by Minu ( 46. If an another disc of same dimensions but M 4 in mass is placed coaxially on first disc gently, then the angular velocity of the system now becomes The moment of inertia (MI) of a disc of radius R and mass M about its central axis is _____. 6k points) rotational motion Dec 25, 2018 · Mass per unit area of disc = M/πR 2. A. of a uniform semicircular disc of mass 'M' and radius 'R' about an axis perpendicular to its pla… Get the answers you need, no… rajdeepbiswas3305 rajdeepbiswas3305 The moment of inertia of a uniform semicircular disc of mass M and radius r about a line perpendicular to the plane of the disc through the centre isA. A thin uniform disc of mass M and radius R has concentric hole of radius r. The MJ. Another disc of the same dimension but of mass M 4 is placed gently on the first disc co-axially. `(MR^(2))/(4) iftheta =0^(@)` B. Step 2: Formula used: Moment of inertia of a uniform circular plate about its axis-I = M R 2 2. Mass of removed portion of disc, Moment of inertia of removed \ portion about an axis passing through centre of disc O and perpendicular to the plane of disc, When portion of disc would not have been removed, the moment of inertia of complete disc about centre O is. , about an axis passing through its centre of mass and perpendicular to its plane is 23) The M. /4. [M. A uniform disc in rolling on horizontal surface at a uniform speed of 6r. The M. 2/5 Mr 2B. Masses of the two semicircular parts are `M` and `3M`. Answer to use I = integral r^2dm to calculate the moment of. A system of solid discs, each of mass M and radius R is shown in figure. Find the moment of inertia of the disc about an axis passing through its centre and perpendicular to its plane. d is the distance between two axis From a uniform circular disc of radius R and mass 9 M, a small disc of radius R 3 is removed as shown in the figure. (Given that moment of inertia of removed portion about the given axis is M R 2 / 2) Q. Its centre of mass is at a distance x from C. (Given that moment of inertia of removed portion about the given axis is M R 2 / 2) The moment of inertia of a thin, uniform, flat semicircular disc of mass m and radius r for rotation about its diameter AB is mr 2 /4. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. of the disc about the same axis is: Medium View solution From a semi-circular disc of mass M and radius R 2. of uniform semicircular wire of mass M and Radius R (as shown in figure) Q. View Solution Q 3 A uniform disk with mass m = 9. Science; Advanced Physics; Advanced Physics questions and answers; use I = integral r^2dm to calculate the moment of inertia of a uniform, solid disk with mass M and radius R for an axis perpendicular to the plane of the disk passing through its center. A circular hole of radius 4 R is made in a thin uniform disc having mass M and radius R, as shown in figure. Using the information given, find the moment of inertia From a uniform circular disc of radius R and mass 9 M, a small disc of radius R 3 is removed as shown in the figure. If the mass of the original uncut disc is M. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O isA. Mass of each disk = M. Taking point A as the origin, the x-axis along AB and the y-axis perpendicular to AB, the centre of mass of the disc is located at co- ordinates (r, 4r/3n). The radius of the disc is 1 decimetre and mass is Find the work done to make it rotate with double the speed. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is A thin disc of mass M and radius R has mass per unit area σ (r) = k r 2, where r is the distance from its centre. There is a smooth groove along the diameter of the disc and two small balls of mass m / 2 each, are placed in it on either side of the centre of the disc as shown in the figure. The time period of small oscillations of remaining portion about O is A disc of radius R, is cut from a disc of radius R 2 as shown in figure. A semi-circular disc of radius ${{R}_{1}}$ is removed as shown in the figure. 7k points) The moment of inertia of a uniform semicircular disc of mass M and radius R about a line perpendicular to the plane of disc and passing through the centre is : From a uniform circular disc of radius R and mass 9 M, a small disc of radius R 3 is removed as shown in the figure. 6. Find the centre of mass of a uniform semicircular disc of radius R. 42 kg and radius R = 1. A thin uniform circular disc of mass M and radius R is rotating with angular velocity Ω in a horizontal plane about an axis passing through its centre and perpendicular to its plane. Q. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is Click here👆to get an answer to your question ️ Find the M. ygwa xwcht udznqgk hvyohx ikhxze sdwt omqqd hlmfq xqoan ubcv