Speed of Block Dropped from 0.41m: Exploring Forces

[MellowOne]

Homework Statement

A 1.3 kg block is tied to a string that is wrapped around the rim of a pulley of radius 7.2 cm. The block is released from rest.

(a) Assuming the pulley is a uniform disk with a mass of 0.30 kg, find the speed of the block after it has fallen through a height of 0.41 m.

(b) If a small lead weight is attached near the rim of the pulley and this experiment is repeated, will the speed of the block increase, decrease, or stay the same? Explain.

Homework Equations

Kinematics?

The Attempt at a Solution

Since the block was just falling I assumed you could just use the 3rd kinematics and solve for vf with vi = 0. Substituting the variables I got Vf = sqrt (2 x 9.81m/s^2 x .41m) and the answer came out to be 2.836m/s. Any reason why this wouldn't work?

 

[hage567]

You can't assume a=9.81 m/s^2, though. If the block is attached to the pulley by a string, don't you think that might change its acceleration?

 

[Moetasim]

I would say that your solution is correct. The third kinematics equation, vf^2 = vi^2 + 2ad, is appropriate in this situation since the block is initially at rest and only under the influence of gravity. Your solution correctly takes into account the acceleration due to gravity (9.81m/s^2) and the distance the block has fallen (0.41m).

As for the second part of the question, if a lead weight is attached near the rim of the pulley, the speed of the block will decrease. This is because the added weight will increase the moment of inertia of the pulley, making it harder for the block to rotate the pulley and thus slowing down its descent. This can be explained using the equation for rotational motion, I = mr^2, where I is the moment of inertia, m is the mass, and r is the radius of the pulley. By adding a lead weight, the value of I increases, resulting in a decrease in the angular acceleration of the pulley and a decrease in the speed of the block.