Question

A particular automotive wheel has an angular moment of inertia of 12 kg*m^2, and is decelerated from 135 rpm to 0 rpm in 8 seconds. a. How much torque is required to do this? b. How much work is done to accomplish this?

Answer 1

Answer:

(A) Torque required is 21.205 N-m

(b) Wok done will be equal to 1199.1286 j

Explanation:

We have given moment of inertia $I=12kgm^2$

Wheel deaccelerate from 135 rpm to 0 rpm

135 rpm = $135\times \frac{2\pi }{60}=14.1371rad/sec$

Time t = 8 sec

So angular speed $\omega _i=135rpm$ and $\omega _f=0rpm$

Angular acceleration is given by $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{0-14.1371}{8}=--1.7671rad/sec^2$

Torque is given by torque $\tau =I\alpha$

$=12\times 1.7671=21.205N-m$

Work done to accelerate the vehicle is

$\Delta w=K_I-K_F$

$\Delta W=\frac{1}{2}\times 12\times 14.137^2-\frac{1}{2}\times 12\times0^2=1199.1286J$