A pulley has the shape of a uniform solid dis | vedclass

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

Question

(D) The moment of inertia $I$ of a uniform solid disc about its central axis is given by:$I = \frac{1}{2} mR^2$Substituting the given values ($m = 2 \

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(D) The moment of inertia $I$ of a uniform solid disc about its central axis is given by:$I = \frac{1}{2} mR^2$Substituting the given values ($m = 2 \ kg$,$R = 0.5 \ m$):$I = \frac{1}{2} 	imes 2 	imes (0.5)^2 = 0.25 \ kg \cdot m^2$The torque $	au$ produced by the force $F = 2.5 \ N$ applied at the rim (radius $R = 0.5 \ m$) is:$	au = F 	imes R = 2.5 	imes 0.5 = 1.25 \ N \cdot m$Using the relation $	au = I \alpha$,we find the angular acceleration $\alpha$:$1.25 = 0.25 	imes \alpha$$\alpha = \frac{1.25}{0.25} = 5 \ rad/s^2$Since the pulley starts from rest $(\omega_0 = 0)$,the angular velocity $\omega$ after $t = 10 \ s$ is:$\omega = \omega_0 + \alpha t$$\omega = 0 + 5 	imes 10 = 50 \ rad/s$

Similar Questions

The angular momentum of a wheel changes from $2L$ to $5L$ in $3$ seconds. The magnitude of the torque acting on the wheel is:

$A$ constant torque of $400 \ Nm$ acts on a wheel having a moment of inertia of $100 \ kg \ m^2$ about its central axis. The angular velocity acquired in $4 \ s$ is ....... $rad \ s^{-1}$.

$A$ wheel is subjected to a torque of $1000 \ N-m$ and rotates with a moment of inertia of $200 \ kg-m^2$ about an axis passing through its center. What will be the angular velocity of the wheel after $3 \ s$?

$A$ 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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module