The rim of a disk with a moment of inertia of 10kg.m^2 and a radius of 0.5m is constantly affected by a force...

Question

The rim of a disk with a moment of inertia of 10kg.m^2 and a radius of 0.5m is constantly affected by a force of magnitude F= 2t+ t^2 (F, in Newton, t in seconds) in the tangential direction.If the disk does not rotate at t=0,

a)Find the disk’s angular acceleration at t=2s,

b)Find the disk’s angular acceleration at t=2s,

c)Find the work done by the force on the disk between t=0 ve t=2s.

d)Find the instantaneous power on the disk by the force at t=1s.

Answer 1

I am assuming no friction or any other form of resistance.

Let

r = 0.5 m be the radius of the disk,

I = 10 kgm² be its moment of inertia,

F(t) = 2t + t² be the tangential force,

τ(t) = rF(t) be the torque,

α(t) be the angular acceleration of the disk,

ω(t) be its angular velocity,

s(t) be the distance travelled by an arbitrarily chosen point on the disk,

W(t) be the work the force has done between 0 and time t,

P(t) be the power exerted on the disk at time t.

a), b)

By Newton's Second Law applied to circular motion

τ(t) = Iα(t)

So

α(t) = τ(t)/I = r( 2t + t²)/I

Substituting actual numbers

α(2) = 0.5(2×2 + 2²)/10 = 0.4 s^-2

c)

Let's first calculate

ω(t) = ∫α(t)dt

= ∫r(2t + t²)/I dt

= r(t² + t³/3)/I + C

The constant C is obtained from the given information that ω(0) = 0, so

C = 0

There are two ways of calculating work -- using Newton's Second Law, and

using the relationship between work and energy.

I will show you both.

Using Newton's Second Law:

By definition of work

F = dW/ds

= dW/dt dt/ds

= dW/dt 1/(ds/dt)

= dW/dt 1/(rω(t))

So

dW/dt = Frω(t) = (2t + t²)r²(t² + t³/3)/I (1)

W = (r²/I) ∫(2t + t²)(t² + t³/3)dt

= (r²/I)(t² + t³/3)²/2 + C

The constant C is calculated from the constraint that at time 0 the force has done 0 work.

So C = 0.

Using Work=Energy relationship:

At time t the disk will have kinetic energy

E(t) = E(0) + Iω²(t)/2

= r²(t² + t³/3)²/2I

Both approaches give us the same formula for work.

Substituting actual numbers

W(2) = 0.5²(2² + 2³/3)²/(2×10) = 5/9 J

d)

By definition of power

P(t) = dW/dt

This is the quantity from (1)

P(t) = (2t + t²)r²(t² + t³/3)/I

Substituting actual numbers

P(1) = (2×1 + 1²)0.5²(1² + 1³/3)/10 = 1/10 W