Two equal and opposite forces are applied tan | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The net torque $	au$ acting on the disc about its centre is the sum of the torques produced by the two forces.Since both forces are applied tange

Vedclass · Accepted Answer

$\frac{4 F}{M R}$

Vedclass · Answer

(C) The net torque $	au$ acting on the disc about its centre is the sum of the torques produced by the two forces.Since both forces are applied tangentially at a distance $R$ from the centre,each force produces a torque of magnitude $	au_i = F 	imes R$.Both forces create a torque in the same rotational direction (counter-clockwise),so the total torque is $	au_{net} = F R + F R = 2 F R$.The moment of inertia of a uniform disc of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I = \frac{1}{2} M R^2$.Using the rotational analogue of Newton's second law,$	au_{net} = I \alpha$,where $\alpha$ is the angular acceleration:$2 F R = (\frac{1}{2} M R^2) \alpha$Solving for $\alpha$:$\alpha = \frac{2 F R}{\frac{1}{2} M R^2} = \frac{4 F}{M R}$.

Two equal and opposite forces are applied tangentially to a uniform disc of mass $M$ and radius $R$ as shown in the figure. If the disc is pivoted at its centre and free to rotate in its plane,the angular acceleration of the disc is ............

Similar Questions

What is the value of the torque acting on an object moving in a circular path with a constant angular speed? Why?

$A$ disc of mass $25 \ kg$ and radius $0.2 \ m$ is rotating at $240 \ r.p.m.$ $A$ retarding torque brings it to rest in $20 \ s$. If the torque is due to a force applied tangentially on the rim of the disc,then the magnitude of the force in newton is

$A$ uniform rod $AB$ of length $l$ and mass $m$ is free to rotate about point $A.$ The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $ml^2/3$,the initial angular acceleration of the rod will be:

$A$ constant torque is applied to a wheel undergoing uniform circular motion,causing its angular momentum to change from $A_0$ to $4A_0$ in $4 \ s$. The magnitude of this torque is .......

$A$ wheel has a moment of inertia $5 \times 10^{-3} \, kg \, m^2$ and is making $20 \, rev/sec$. The torque needed to stop it in $10 \, sec$ is $............. \times 10^{-2} \, N-m$. (in $\pi$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module