# Pulley Problem: Speed after falling a known distance

A 1.2·kg block is tied to a string that is wrapped around the rim of a pulley of radius 7·cm. The block is released from rest.

Assuming the pulley is a uniform disk with a mass of 0.37·kg, find the speed of the block after it has fallen through a height of 0.43·m.

2. F=mrω^2 and possibly F=mg and v=r*omega

3. I wasn't sure how to do this problem at all. I looked around online for help and got a little bit but I'm still unsure. I used F=mg to determine the F then used that to determine ω^2. Is that correct? Obviously there would be more to the problem but this is as far as I got and I'm not even sure if it's correct.

So I got F=11.76 by multiplying 1.2kg * 9.8m/s^2 which I then put in the equation for 11.76=(.37kg)(.07m)w^2 and then got 21.3 for omega. This seems high and is probably due to the conversion from CM to M which I'm unsure is correct. Assuming it is I moved on and put that omega value 21.3 into v=wr for v=21.3*.07 and got 1.49 which is incorrect.

Last edited:

Related Introductory Physics Homework Help News on Phys.org
Gold Member
That depends on the friction of the pulley.

That depends on the friction of the pulley.
I believe it's assumed frictionless because friction isn't mentioned

Gold Member
Conversion from meters to cm is wrong.(thats all I know) 7/1000 is 0.007m

Conversion from meters to cm is wrong.(thats all I know) 7/1000 is 0.007m
Isn't a centimeter 1/100th of meter?

Gold Member
Ah.Sorry,I was mistaken.You are right.

Gold Member
It seems you didn't read this part(find the speed of the block after it has fallen through a height of 0.43·m. Hope it helps.

haruspex
Homework Helper
Gold Member
You need to consider the pulley and the mass separately. What they have in common are the tension in the string and the acceleration. Draw a FBD for each and write out the equations (ƩF=ma etc.)