Rotational Dynamics

1. Name: ____________________
Rotational Motion Problem Set C
C. Torque
1. A bucket filled with water has a mass of 54 kg and is attached to a rope that is wound
around a 5.0 cm radius cylinder. What magnitude of torque does the bucket produce
around the center of the cylinder?
Since the bucket is hanging from the edge of the cylinder, the weight of the bucket is
the force applied to rotate the cylinder. So torque would be from force x radius
τ=Fxr
F =mg = 54kg(9.8m/s/s) = 529.2 N
τ=529.2N x 0.05m = 26.46 Nm
2. Determine the net torque about the axle of the wheel shown below.
First, establish that there are 2 directions of rotation
Clockwise, cw and counterclockwise, ccw
Clockwise is negative and ccw is positive directions
You can see from the picture that the forces are applied at the edge of 2 wheels.
This means that torque is applied and can be found by taking force x radius
τ=Fxr for 15 N τ=15 N x 0.32 m in a cw or negative direction = -4.8 Nm
for 10 N τ=10N x 0.2 m in a ccw or + direction = +2Nm
Net force is the sum of the 2 torques = -4.8 nm + 2 Nm = -2.8 Nm
3. The arm of a crane is 15.0 m long and makes an angle of 20.0 with the horizontal.
Assume that the maximum load for the crane is limited by the amount of torque the
load produces around the base of the arm.
a) What is the magnitude of the maximum torque the carne can withstand if the
maximum load is 450 N?
b) What is the magnitude of the maximum load for this crane at an angle of 40.0?
To calculate torque, you need the force and lever arm to be perpendicular to each other. So, we
must find a component of either the lever arm or the force to make each perpendicular to each
other.
Easiest route would be to find the x comp of 15m or 15mcos20 = 14.1 m
20. cm
32 cm
15.0 N
10.0 N
15m
450 N

2. Torque is then τ=Fxr = 450 N x 14.1m = 6342.9 Nm
b. For this part, we know what the max torque is from part a. So now it is a matter of figuring
out the force that the max torque can provide
τ=Fxr same set up so now we need the x comp of 15 m at 40 degrees
τ=Fxr F = τ/r = 6342.9 N /15cos40 = 6342.9Nm/11.5m = 551.6 N
4. Find the torque exerted on the fishing pole by a fish pulling with a force of 100 N as
shown below.
Based on the picture, we see that the force and lever arm are not perpendicular so we need to
find a component of one that is perpendicular to the other.
Find the component of the string tension, that is vertical. This would be perpendicular to the
pole.
Sin20 = Fy/100N Fy = 100Nsin20 = 34.2 N
τ=Fxr 34.2Nx2m = 68.4 Nm
1) 26.5 mN 2) -2.80 mN
3) a. 6340 mN
b. 552 N
4) 68.4 mN