Question

A 50n brick is suspended by a light string from a 30kg pulley which may be considered a solid disk with radius 2.0m. the brick is released from rest and falls to the floor below as the pulley rotates. it takes 4 seconds for the brick to hit the floor. i) what is the tension in newtons in the string well the brick is falling? ii) what is the magnitude of the angular momentum in kg*m^2/s of the pulley at the instant the brick hits the floor?

Answer 1

Brick is held at a position which is at height 2 m from the floor

Now it is released from rest and hit the floor after t = 4 s

Now the acceleration of the brick is given by

$y = v_i* t + 0.5 at^2$

$2 = 0 + 0.5 * a * 4^2$

$a = 0.25 m/s^2$

a)

Now in order to find the tension in the string

we can use Newton's law

$F_{net} = ma$

$mg - T = ma$

$50 - T = \frac{50}{9.8}*0.25$

$T = 48.72 N$

part b)

Now for the pulley

moment of inertia= $\frac{1}{2}mr^2$

m = 30 kg

R = 2 m

I = $\frac{1}{2}*30*2^2$

I = 60 kg m^2

Now the angular speed just before brick collide with the floor

$w = \frac{v}{r}[\tex]here we have[tex]v = v_i + a* t$

$v = 0 + 0.25 * 4$

v = 1 m/s

Now we will have

L = angular momentum = I w = $I*\frac{v}{R}$

L = 60 *$\frac{1}{2}$

L = 30 kg m^2/s