A circular hoop of radius 50 cm and mass 1 | vedclass

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

Question

(C) The initial angular momentum of the hoop about the point of contact is $L_i = I_{cm} \omega_0 + m r^2 \omega_0 = m r^2 \omega_0 + m r^2 \omega_0 =

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) The initial angular momentum of the hoop about the point of contact is $L_i = I_{cm} \omega_0 + m r^2 \omega_0 = m r^2 \omega_0 + m r^2 \omega_0 = 2 m r^2 \omega_0$. However, considering the angular momentum about the contact point $P$ at the instant it is placed: $L_i = I_P \omega_0 = (I_{cm} + m r^2) \omega_0 = (m r^2 + m r^2) \omega_0 = 2 m r^2 \omega_0$.When the hoop stops slipping, it rolls without slipping, so $v = r \omega$. The final angular momentum about the contact point $P$ is $L_f = I_{cm} \omega + m r v = m r^2 (v/r) + m r v = m r v + m r v = 2 m r v$.Since the frictional force acts through the contact point, the net torque about the contact point is zero. Thus, angular momentum is conserved: $L_i = L_f$.$2 m r^2 \omega_0 = 2 m r v$.Dividing both sides by $2 m r$, we get $v = r \omega_0$.Wait, re-evaluating: $L_i = I_{cm} \omega_0 = m r^2 \omega_0$. The angular momentum about the contact point is $L_i = I_{cm} \omega_0 = m r^2 \omega_0$.$L_f = I_{cm} \omega + m r v = m r^2 (v/r) + m r v = 2 m r v$.Equating $L_i = L_f$: $m r^2 \omega_0 = 2 m r v$.Therefore, $v / \omega_0 = r / 2 = 50 \ cm / 2 = 25 \ cm$.

Similar Questions

$A$ uniform non-deformable cylinder of mass $m$ and radius $R$ is rolling without slipping on a horizontal rough surface. The force of friction is

Consider a point $P$ at the contact point of a wheel on the ground,which rolls on the ground without slipping. Find the displacement of point $P$ when the wheel completes half of a rotation (given the radius of the wheel is $1 \ m$).

In a bicycle,the radius of the rear wheel is twice the radius of the front wheel. If $r_F$ and $r_r$ are the radii,and $v_F$ and $v_r$ are the speeds of the topmost points of the wheels respectively,then:

$A$ disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in the figure. At any instant,for the lowermost point of the disc -

Find the displacement of the point of contact with the ground when a ring of radius $R$ completes a half-rotation.

Vedclass Test Series

Exam Paper Generator

Online Exam Module