Question

You are designing a lathe motor, part of which consists of a uniform cylinder whose mass is 130 kg and whose radius is 0.40 m that is mounted so that it turns without friction on its axis, which is fixed. The cylinder is driven by a belt that wraps around its perimeter and exerts a constant torque. At t = 0, the cylinder's angular velocity is zero. At t = 25 s, its angular speed is 450 rev/min.

(a) What is the magnitude of the cylinder's angular momentum at t = 25 s?
(b) At what rate is the angular momentum increasing?
(c) What is the magnitude of the torque acting on the cylinder?
(d) What is the magnitude of the frictional force acting on the rim of the cylinder?

Answer 1

a) 490.152 kgm²/s

b) 19.61

c) 19.61

d) 49.025 N

Step-by-step explanation:

a)

Angular momentum, L = Iw

Iw = moment of inertia * angular velocity

I = 1/2 MR² for a uniform cylinder

w = 450 rpm. Converting to rad/s, we have

450 * (2 * 3.142/60) = 47.13 rad/sec

I = 1/2 * 130 * 0.4²

I = 10.4 kgm²

Now, proceeding to find L,

L = Iw

L = 10.4 * 47.13

L = 490.152 kgm²/s

b.)

rate = ΔL / Δt

Rate = 490.152 / 25sec

Rate = 19.61

c.)

T = dL/dt which is essentially the same as 19.61

d.)

T = Fr, making F subject of formula, we have

F = T/r

F = 19.61 / 0.4 = 49.025 N