An annular disk of mass M,inner radius a and | vedclass

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

Question

(D) For an impulse $J_0$ applied at height $h$ from the center:$1$. Linear impulse equation: $J_0 = Mv \implies v = \frac{J_0}{M}$$2$. Angular impulse

Vedclass · Accepted Answer

$A, B, C, D$

Vedclass · Answer

(D) For an impulse $J_0$ applied at height $h$ from the center:$1$. Linear impulse equation: $J_0 = Mv \implies v = \frac{J_0}{M}$$2$. Angular impulse equation about the center of mass: $J_0 h = I_c \omega \implies \omega = \frac{J_0 h}{I_c}$For rolling without slipping,$v = \omega b$,where $b$ is the outer radius.Substituting $v$ and $\omega$: $\frac{J_0}{M} = \frac{J_0 h_m}{I_c} b \implies h_m = \frac{I_c}{Mb}$For an annular disk,$I_c = \frac{1}{2} M(a^2 + b^2)$.Thus,$h_m = \frac{\frac{1}{2} M(a^2 + b^2)}{Mb} = \frac{a^2 + b^2}{2b}$.$(A)$ If $a ightarrow 0$,$h_m = \frac{b^2}{2b} = \frac{b}{2}$. Statement $(A)$ is correct.$(B)$ If $a ightarrow b$,$h_m = \frac{b^2 + b^2}{2b} = \frac{2b^2}{2b} = b$. Statement $(B)$ is correct.$(C)$ $\omega = \frac{J_0 h_m}{I_c} = \frac{J_0}{I_c} \cdot \frac{I_c}{Mb} = \frac{J_0}{Mb}$. Since $M$ and $b$ are constants,$\omega$ does not depend on $a$. Statement $(C)$ is correct.$(D)$ If $\mu = 0$ and $h = 0$,the impulse passes through the center of mass,so no torque is produced $(	au = 0)$. Thus,$\omega = 0$ and the disk only translates. It slides without rolling. Statement $(D)$ is correct.

Similar Questions

$A$ meter stick is held vertically with one end on the floor and is allowed to fall. The speed of the other end when it hits the floor,assuming that the end at the floor does not slip,is ......... $m/s$ $(g = 9.8 \, m/s^2)$.

$A$ thin uniform rod of length $l$ and mass $m$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of:

$A$ uniform thin rod of length $2L$ and mass $m$ lies on a horizontal table. $A$ horizontal impulse $J$ is given to the rod at one end. There is no friction. The total kinetic energy of the rod just after the impulse will be

The angular momentum of a particle performing uniform circular motion is $L$. If the kinetic energy of the particle is doubled and the frequency is halved,then the new angular momentum becomes:

Vedclass Test Series

Exam Paper Generator

Online Exam Module