1. Rotational Dynamics

2. A resultant force F produces negative acceleration a for a mass m.
Important Analogies
For many problems involving rotation, there is an analogy to be drawn from linear motion. R
4 kg w t
wo = 50 rad/s t = 40 N m m v F I
A resultant torque t produces angular acceleration a of a disk with rotational inertia I.
A resultant force F produces negative acceleration a for a mass m.

3. What is Torque? A torque is an action that causes an object to rotate.
Torque is not the same thing as force.
Torque is created when the line of action of a force does not pass through the center of rotation.
The line of action is an imaginary line that follows the direction of a force and passes though its point of application.

4. t = r x Fsinθ What is Torque? Lever arm length (m) Torque (N.m)
Force (N)
What about direction?
CCW CW

5. Example: Rank the Torques
τa > τb > τc

6. Newton’s 2nd Law for Rotation Example
t = Ia
How many revolutions are required to stop the disk? (I = ½ mR2) R
4 kg w F
wo = 50 rad/s
R = 0.20 m
F = 40 N
FR = (½mR2)a
2aq = wf2 - wo2
a = 100 rad/s2
q = 12.5 rad = 1.99 rev

7. T mg - T = ma mg - = ma ½Ma a = 3.92 m/s2
Example 3: What is the linear acceleration of the falling 2-kg mass? (I = ½ MR2)
R = 50 cm
6 kg
m = 2 kg
a = ? M
Apply Newton’s 2nd law to rotating disk:
t = Ia
TR = (½MR2)a
a = aR; a =
but aR
T = ½MRa
R = 50 cm
6 kg
2 kg +a T mg
T = ½MR( ) ; aR
and
T = ½Ma T
Apply Newton’s 2nd law to falling mass:
mg - T = ma
mg = ma
½Ma
(2 kg)(9.8 m/s2) - ½(6 kg) a = (2 kg) a
19.6 N - (3 kg) a = (2 kg) a
a = 3.92 m/s2

8. Torque & Static Equilibrium: Example Problem
A boy and his cat sit on a seesaw. The cat has a mass of 4 kg and sits 2 m from the center of rotation. If the boy has a mass of 50 kg, where should he sit so that the see-saw will balance?
1) You are asked to find the boy’s lever arm.
2) You are given the two masses and the cat’s lever arm.
3) Torque, τ = rF weight: F = mg
In equilibrium the net torque must be zero.
4) Solve: For the cat,
τ = (2 m)(4 kg)(9.8 N/kg) = N-m
For the boy:
τ = (d)(50 kg)(9.8 N/kg) = d
For rotational equilibrium, the net torque must be zero.
d = 0
d = 0.16 m
The boy must sit quite close (16 cm) to the center.

9. Calculate Using Static Equilibrium
A boy and his cat sit on a seesaw. The cat has a mass of 4 kg and sits 2 m from the center of rotation. If the boy has a mass of 50 kg, where should he sit so that the see-saw will balance?
1) What are you trying to find? The boy’s lever arm (r).
2) You are given the two masses and the cat’s lever arm.
3) Torque, τ = rF weight: F = mg
In equilibrium the net torque must be zero.
4) Solve:
For the cat,
τ = (2 m)(4 kg)(9.8 N/kg) = N-m
For the boy:
τ = (d)(50 kg)(9.8 N/kg) = d
For rotational equilibrium, the net torque must be zero. So the CW torques must equal the CCW torques.
d = 0 OR 78.4 = 490d
d = 0.16 m, The boy must sit quite close (16 cm) to the center.
1) You are asked to find the boy’s lever arm.
2) You are given the two masses and the cat’s lever arm.
3) Torque, τ = rF weight: F = mg
In equilibrium the net torque must be zero.
4) Solve: For the cat,
τ = (2 m)(4 kg)(9.8 N/kg) = N-m
For the boy:
τ = (d)(50 kg)(9.8 N/kg) = d
For rotational equilibrium, the net torque must be zero.
d = 0
d = 0.16 m
The boy must sit quite close (16 cm) to the center.