Question

The lightweight wheel on a road bike has a moment of inertia of 0.097 kg⋅m2. A mechanic, checking the alignment of the wheel, gives it a quick spin; it completes 5 rotations in 2.4 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 1.4 s.


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

Answer 1

Answer:

The magnitude of the friction force on the disk is 12.76 N.

Explanation:

Given that,

Moment of inertia = 0.097 kg m²

Number of rotation = 5

Time = 2.4 s

Distance = 7.1 cm

Time = 1.4 s

We need to calculate the angular velocity

Using formula of angular velocity

$\omega=\dfrac{n2\pi}{t}$

Put the value into the formula

$\omega=\dfrac{5\times2\pi}{2.4}$

$\omega=13.08\ rad/s$

We need to calculate the magnitude of the friction force on the disk

Using formula of torque

$\tau=I\alpha$

$F\cdot r=I\alpha$

Put the value into the formula

$F\times7.1\times10^{-2}=0.097\times\dfrac{13.08}{1.4}$

$F\times7.1\times10^{-2}\times1.4=0.097\times13.08$

$F\times0.0994=1.26876$

$F=\dfrac{1.26876}{0.0994}$

$F=12.76\ N$

Hence, The magnitude of the friction force on the disk is 12.76 N.

Answer 2

Answer:

$F=12.774\ N$

Explanation:

Given:

moment of inertia of the wheel, $I=0.097\ kg.m^2$

no. of rotation of the wheel, $n=5$

time taken to complete the above said rotations, $t_r=2.4\ s$

distance of application of force by the brake pads, $r_b=0.071\ m$

time taken to stop, $t_s=1.4\ s$

Now the angular velocity of the wheel:

$\omega=\frac{2\pi.n}{t_2}$

$\omega=\frac{2\pi\times 5}{2.4}$

$\omega=13.09\ rad.s^{-1}$

According the Newton's second law of motion:

$\tau=I.\alpha\\F.r_b=I.\frac{\omega}{t_s}$

$F\times r_b=0.097\times \frac{13.09}{1.4}$

$F\times 0.071=0.9069$

$F=12.774\ N$