Question

A rough strip of negligible mass and length 10cm between two wheels pivoted at their centers is pulled out with a constant force of 20N. It sets the two wheels rotating, a) What work is done in pulling the strip out? b) What must the rotational energy of each wheel be? c) What is the total angular momentum of the wheels after they are rotating? d) What total angle has each of the wheels rotated through when the 10cm strip just leaves if their radii are 1cm? e) What is the angular velocity of each wheel after the strip has been pulled out?

Answer 1

The moment the strip starts, a constant torque starts acting on the wheels whose magnitude can be given by

τ = Fr = (20)(1/100)= 0.2 N-m

it acts till the whole strip of length 10 cm is pulled out that means it acts till the wheels rotate through ϴ = l/r = 10/1 = 10 rad

hence the work done = 2(τϴ)= 2(0.2)(10) = 4 J (the factor of 2 comes due to two wheels)

now as initially the wheels were at rest and now they are rotating hence the change in kinetic energy = amount of work done

2(Iω²/2) = 4

Iω² = 4 where I = mr²/2

Now

Part a

The work done in pulling the strip = 4 J

Part b

The rotational kinetic energy of each wheel, E = Iω²/2 = 2 J

Part c

Angular momentum = (2IE)^1/2 (using relation E= L²/2I)

Part d

Clearly the required angle = 10 rad

Part e

As

Iω² = 4

ω= (4/I)^1/2

(note we need mass or moment of inertia of the wheels in order to complete the part c and e)