A disc has mass M and radius R. How much tang | vedclass

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

Question

(B) The moment of inertia of a disc about its central axis is $I = \frac{1}{2} MR^2$.Using the equation of rotational motion,$\omega = \omega_0 + \alp

Vedclass · Accepted Answer

$\frac{MR \omega}{2 t}$

Vedclass · Answer

(B) The moment of inertia of a disc about its central axis is $I = \frac{1}{2} MR^2$.Using the equation of rotational motion,$\omega = \omega_0 + \alpha t$. Since the disc starts from rest,$\omega_0 = 0$,so $\alpha = \frac{\omega}{t}$.The torque $	au$ required is given by $	au = I \alpha = (\frac{1}{2} MR^2) 	imes (\frac{\omega}{t}) = \frac{MR^2 \omega}{2t}$.Since the force $F$ is applied tangentially at the rim,the torque is $	au = F 	imes R$.Equating the two expressions for torque: $F 	imes R = \frac{MR^2 \omega}{2t}$.Therefore,the tangential force is $F = \frac{MR \omega}{2t}$.

$A$ disc has mass $M$ and radius $R$. How much tangential force should be applied to the rim of the disc,so as to rotate with angular velocity $\omega$ in time $t$?

Similar Questions

$DCMU$ is an inhibitor of

Which of the following is a non-reducing sugar?

If $A=\begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix}$ and $f(t)=t^2-3t+7$,then $f(A)+\begin{bmatrix} 3 & 6 \\ -12 & -9 \end{bmatrix}$ is equal to

The power dissipated in the circuit shown in the figure is $30 \, W$. The value of $R$ is ............. $\Omega$.

$A$ charge '$q$' is spread uniformly over an isolated ring of radius '$R$'. The ring is rotated about its natural axis with angular speed $\omega$. The magnetic dipole moment of the ring is

Vedclass Test Series

Exam Paper Generator

Online Exam Module