Question

The lightweight wheel on a road bike has a moment of inertia of 0.097 kg⋅ m 2 . A mechanic, checking the alignment of the wheel, gives it a quick spin; it completes 5 rotations in 2 s . To bring the wheel to rest, the mechanic gently applies the disk brakes, which squeeze pads against a metal disk connected to the wheel. The pads touch the disk 7.1 cm from the axle, and the wheel slows down and stops in 1.5 s .

A) what is the initial and final angular velocity of the wheel, before application of brakes?

B) What is the angular acceleration just before the wheel stopped?

C) What is the torques due to this angular acceleration?

D) What is the amount of force that excreted such a torque?

E) What is the magnitude of the friction force on the disk?

Answer 1

Moment of inertia of wheel=0.097 kg-m²

a)initially, the wheel completes 5 rotations in 2s,

So, initial angular velocity=(angular displacement/time)=5*2/2=5 rad/sec=15.71 rad/sec

Finally, the wheel stops. So,final angular velocity=0 rad/sec

b)When friction is applied the wheel comes to rest in 1.5 s

angular acceleration=(final angular velocity-initial angular velocity)/time=(0-5)/1.5=-10.47 rad/s²

The negative sign denotes deceleration with magnitude 10.47 rad/s²

c)Torque=moment of inertia*angular acceleration

Moment of inertia=0.097 kg-m² , Angular acceleration=-10.47 rad/s²

So,torque=-1.01559 N-m= -1.02 N-m

So, torque has magnitude 1.02 N-m and its direction is opposite to that of angular velocity.

d)Magnitude of torque=r*F, where r is the magnitude of position vector (taking the point about which torque is to be calculated as origin) where force is being applied and F is magnitude of force.

Here r=7.1 cm=0.071 m

Also magnitude of torque=1.02 N-m

So, 1.02=0.071*F=>F=1.02/0.071=14.37 N

So, the amount of force that exerted this torque is 14.37 N

e)Magnitude of friction force=amount of force that exerted the torque=14.37 N