A wheel rotates about its geometric axis at a | vedclass

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

Question

(C) Initial angular velocity $\omega_i = 60 \ rpm = \frac{60 	imes 2\pi}{60} = 2\pi \ rad \ s^{-1}$.Final angular velocity $\omega_f = 0 \ rad \ s^{-

Vedclass · Accepted Answer

$\frac{\pi}{15} \ Nm$

Vedclass · Answer

(C) Initial angular velocity $\omega_i = 60 \ rpm = \frac{60 	imes 2\pi}{60} = 2\pi \ rad \ s^{-1}$.Final angular velocity $\omega_f = 0 \ rad \ s^{-1}$.Time taken $t = 1 \ minute = 60 \ s$.Moment of inertia $I = 2 \ kg \ m^2$.The angular acceleration is given by $\alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - 2\pi}{60} = -\frac{\pi}{30} \ rad \ s^{-2}$.The required torque is $	au = I\alpha = 2 	imes \left( -\frac{\pi}{30} ight) = -\frac{\pi}{15} \ Nm$.The magnitude of the torque required is $\frac{\pi}{15} \ Nm$.

$A$ wheel rotates about its geometric axis at a speed of $60 \ rpm$. If the moment of inertia of the wheel about this axis is $2 \ kg \ m^2$,what torque is required to stop its rotation in one minute?

Similar Questions

$A$ 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 is:

$A$ flywheel attains a speed of $540 \, r.p.m.$ in $6 \, s$. Its angular acceleration is:

If a constant torque acts on an object,then the object:

Obtain the relation between torque and moment of inertia.

$A$ wheel of mass $20 \,kg$ and radius $30 \,cm$ is rotating at an angular speed of $80 \,rev/min$ when the motor is turned off. Neglecting the friction at the axis, calculate the force that must be applied tangentially to the wheel to bring it to rest in $5 \,rev$. (in $\pi \,N$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module