TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Given that, mass of wheel, m=20kg
Radius, R=30cm=30×10−2m
‌ Initial angular speed, ‌ω0‌‌=80rpm=‌

80×2π

60

‌‌=‌

8π

3

rad∕s
Angular displacement, θ=2π×( number of revolution) =2π×5=10πrad
Final angular speed, ω=0
By equations of rotational kinematics.
Angular acceleration, α=‌

ω2−ω02

2θ

Substituting the given values, we get
α‌‌=‌

0−( 8π 3 )2

2×10π

rad∕s2
‌‌=−‌

16π

45

rad∕s2
Now, the tangential force will act opposite to the direction of rotation of wheel which will provide necessary retarding torque.
∴‌τ=Iα=FR
⇒‌F=‌

Iα

R

=‌

‌ 1 2 mR2α

R

=‌

1

2

mRα
Substituting the values, we get
F=‌

1

2

×20×30×10−2×‌

16π

45

N=1.06πN