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 1.6 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 1.3 s.What is the magnitude of the friction force on the disk?

Answer 1

The magnitude of the friction force is 20.63N

Step-by-step explanation:

Given that,

Moment of inertia, I is 0.097 kg m²

Number of rotation, n is 5

completes 5 rotations, Time, t= 1.6 s

Distance is 7.1 cm = 0.071 m

Time, t = 1.3 s

Angular velocity is

`\omega = (n 2 \pi)/(t)`

`\omega = (5 * 2 \pi)/(1.6) \\\\\omega = 19.63 rad/s`

The magnitude of the friction force on the disk with the help of torque formula

τ= Iα

I = moment of inertia (kg∙m2)

α = angular acceleration (radians/s2)

Angular acceleration (α) can be defined as angular velocity (ω) divided by acceleration time (ts)

F.r =Iα-----(α= ω/ts)

F x 0.071 = 0.097 (19.625/1.3)

F x 0.071 x 1.3 = 0.097 x 19.63

F= 20.63N

Thus, the magnitude of the friction force is 20.63N

Answer 2

20.62N

Step-by-step explanation:

According to the given data:

Moment of inertia 'I'= 0.097 kg m²

Number of rotation 'n'= 5

completes 5 rotations, Time 't'= 1.6 s

Distance = 7.1 cm=> 0.071 m

wheel slows down and stops, Time 'ts'= 1.3 s

First let calculate the angular velocity i.e

ω= (n2π)/t

ω= (5 x 2π)/1.6

ω= 19.625 rad/s

next is to calculate the the magnitude of the friction force on the disk with the help of torque formula i.e

τ= Iα

I = moment of inertia (kg∙m2)

α = angular acceleration (radians/s2)

Angular acceleration (α) can be defined as angular velocity (ω) divided by acceleration time (ts)

F.r =Iα----->(α= ω/ts)

F x 0.071 = 0.097 (19.625/1.3)

F= 20.62N

Therefore, the magnitude of the friction force on the disk is 20.62N