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# AP Physics 1 review of Torque and Angular momentum

In this video, David quickly explains each torque and angular concept and does a sample question for each one. Created by David SantoPietro.

## Want to join the conversation?

• At ,why does the clay have an ANGULAR momentum before the collision? I thought it was moving translationally?Thanks
• Think about it this way: Once the translationally moving mass comes into contact with the rod and sticks to it, its translational momentum will become angular momentum with the same magnitude, so what you are calculating is not it's angular momentum while flying in a straight line, but it's "potential angular momentum" that it would have if it started moving in a circle.
• At , why did David use the rotational inertia of the clay sphere to be ML^2 when the formula of the rotational inertia of a sphere is (2/5)(M)(L^2)?
• So does rotational kinetic energy affect mechanical energy or just translational kinetic energy?
(1 vote)
• Mechanical energy is the sum of KINETIC and gravitational energy so I believe that rotational KINETIC energy would affect mechanical energy.
• What is the right hand rule?
• Given that you're on the video for rotational motion, the right-hand rule (for rotational motion) applies to vector quantities (angular velocity, angular acceleration, torque, etc.); it basically states that if you curl your right hand in the direction of the rotation, the vector's direction is the direction of your thumb.
(1 vote)
• Can you please specify the differences/relationships between angular acceleration, tangential acceleration, radial acceleration and centripetal acceleration? I keep mixing up the appropriate equations/units. I feel that is because I do not fully understand the subtleties of these terms.
(1 vote)
• Angular acceleration α = dω/dt = d²θ/dt² [rads/sec²]
Angular velocity ω = dθ/dt [rads/sec]
Tangential acceleration a = rα [meters/sec²]
Tangential velocity v = rω [meters/sec]
Tangential displacement s = rθ [meters]
Centripetal acceleration (same thing as radial acc.) a = v²/r = rω² [meters/sec²]

Note that radians are a unitless unit in that they don't have any specific dimension. This is because the radian is defined as the ratio of the circle arc length/radius, so [distance/distance] is unitless.
• At , shouldn't we use the formula of I=mr^2 to culcuate the inertia of those 2 objects? ( since B is a hoop, we can use that formula) If we use that formula, then the answer will be C.
• Your time stamp doesn't direct to the problem you're asking about. Could you recheck it? Or are you referring to a different video altogether?
(1 vote)
• At , shouldn't the torque with 1 newton of force be positive since it is counter-clock wise and the torque with the 4 N force be negative since it is pushing clockwise? The final answer would be the same because the answer asks for the magnitude, but just wondering.
(1 vote)
• (v^2)/r is centripetal acceleration and so is (omega)^2(r). when exactly do we use the latter one? I was told you only use that when the velocity is changing
(1 vote)
• Actually, they are equivalent.

recall: v= r omega

You would simply use them when the data or analysis best suits.

The velocity (strictly speaking) is always changing because the direction is changing

if the speed is changing then you would need to take into account the tangential acceleration (alpha usually) which is different to the centripetal acceleration

ok??
(1 vote)
• I don't understand how the problem was approached for . The sign for length (L) and the sign for angular momentum (L) are the same and are treated as the same thing. Why? Is angular momentum equal to the distance between the axis of rotation and the collision point?
(1 vote)
• No. Notice that when he refers to angular momentum he writes Li, not just L. The i is for inertial. The problem stipulates that the length of the rod is L. It's confusing since we often use L for angular momentum. Here he tried to resolve that confusion by using Li but that was not a good way to do it because the i is hard to even notice. He should have just given the angular momentum a completely different variable name for this problem.
(1 vote)
• At , it says that the gravitational potential energy can be determined using Gm1m2/d. I thought that the formula was Gm1m2/d^2... Or is gravitational attraction different from gravitational potential energy?
(1 vote)
• gravitational force is not the same as gravitation potential energy

Just like mg is not the same as mgh
(1 vote)