You are here:

Physics/rotation of a rigid body


rotating bodies on a pulley
rotating bodies on a p  
The two blocks in the figure(Figure 1) are connected by a massless rope that passes over a pulley. The pulley is 15 cm in diameter and has a mass of 2.6 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.51 Nm. If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

Hello Billy,

The data regarding the pulley in your write-up disagrees with the data on the figure. I'll use variable names. It should be easy enough once you decide which set of data to use. The figure implies that the pulley is more complex than just a disk, but there's no details, so I'll have to treat it as a disk, so its rotational inertia is
I = (1/2)*mp*R^2

The mass m2 will fall, so its weight is greater than tension T2. The polarities related related to m2 falling and the pulley rotating counterclockwise will be considered positive. The rate of m2's acceleration is given by
m2*g - T2 = m2*a ----------- #1
Also we can write
m1*g -T1 = m1*a ------------ #2

The net torque on the pulley is the sum of Tfric, and the torques due to tensions T1, and T2.
torque due to T1 is T1*R
torque due to T2 is -T2*R
Tfric = -0.50 N.m
Tnet = T2*R - T1*R - Tfric
(Tfric is given as a torque so it is not multiplied by R.)

The angular version of Newton's 2nd Law is
Tnet = I*alpha
We get
T2*R - T1*R - Tfric = (1/2)*mp*R^2*alpha ------------ #3

The relationship between angular acceleration, alpha, and linear acceleration, a, is
a = alpha*R
If we divide both sides of the equation labeled #3 by R, plug a in for alpha*R, and do some simplifying, we have
T2 - T1 - Tfric/R = mp*a/2 ----------------- #4

Now let's apply our simultaneous equation skills to equations 1
m2*g - T2 = m2*a
and 4
T2 - T1 - Tfric/R = mp*a/2
Solve #1 for T2:
T2 = m2*(g-a)
Plug that into #3
m2*(g-a) -T1 -Tfric/R = mp*a/2
then T1 = m2*(g-a) - Tfric/R - mp*a/2 -------- #5

Now let's include equation 2
m1*g -T1 = m1*a
Solve for T1:
T1 = m1*(g-a)
Plug that into #5
m1*(g-a) = m2*(g-a) - Tfric/R - mp*a/2

Since I'm not plugging any data in, that looks nasty. But you know everything except a. So solve for a. That's the acceleration of mass m1. Use the kinematic formula
y = Vi*t + (1/2)*a*t^2
where y = 1 m, Vi = 0, and you just found a in the previous step.
(Please, check my algebra. There were too many steps to be sure I got every one right.)

I hope this helps,


All Answers

Answers by Expert:

Ask Experts


Steve Johnson


I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.


I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.

BS Physics, North Dakota State University
MS Electrical Engineering, North Dakota State University

©2017 All rights reserved.