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Peter Apps
Kashvi Panjolia
Peter Apps
Kashvi Panjolia
Angular momentum is the rotational equivalent of linear momentum. It is represented by the equation L=I𝜔 where L is the angular momentum, I is the moment of inertia of the object, and 𝜔 is the angular velocity of the object. In addition, the angular momentum of an object moving in a straight line relative to a fixed point can be found by multiplying its linear momentum (p) by the perpendicular distance (r) to the point as shown in the picture.
🎥Watch: Minutephysics - What is Angular Momentum?
A student will slide a disk of mass:
The student wants the rod-disk system to end up with as much angular speed as possible.
CORRECT ANSWER: To the right of C
REASONING: To make the largest angular speed, the rod needs to get the greatest angular momentum possible (L=I𝜔). To get the largest angular momentum, the rod must be given the greatest possible torque. Since torque depends on the moment of inertia and the angular acceleration (𝜏=Iα), and the moment of inertia is calculated using I=cMR^2 (where c is a constant depending on the object), hitting the rod at the greatest radius will produce the greatest torque. Because the pivot is on the left side of C, we want to hit on the right side of C to maximize the radius so the moment of inertia increases, causing the net torque to increase, which will increase the angular speed.
ALTERNATE EXPLANATION: To create the largest angular speed, the rod needs to get the greatest angular momentum possible (L=I𝜔). To do this, we want the disk to transfer the greatest amount of momentum to the rod. Since angular momentum is conserved in the rod-disk system (no outside torques), this means that we want the initial angular momentum of the disk to be as large as possible. The angular momentum of the disk is equal to L=mvr where r is the distance from the pivot because the disk is moving in a straight-line motion relative to the pivot point of the rod.
To make:
(This explanation uses the law of conservation of momentum which will be covered in the next section.)
🎥Watch: AP Physics 1 - Unit 7 Streams
1. A spinning disk has a mass of 0.5 kg and a radius of 0.2 m. The disk is spinning at an angular velocity of 2 rad/s. What is the angular momentum of the disk? (I = 1/2MR^2 for the disk)
A) 0.02 kgm^2/s B) 0.2 kgm^2/s C) 0.4 kgm^2/s D) 0.8 kgm^2/s
The solution is: L = Iω = (1/2) mr^2ω = (1/2)*0.5(*0.2^2)2 = 0.02 kgm^2/s
The answer is A) 0.02 kgm^2/s
2. A 0.4 kg ball is attached to a string of length 0.5 m and is rotating in a circle with an angular velocity of 3 rad/s. What is the angular momentum of the ball? (I= 2/5MR^2 for the ball)
A) 0.3 kgm^2/s B) 0.12 kgm^2/s C) 0.9 kgm^2/s D) 1.2 kgm^2/s
The solution is:
L = Iω = (2/5) mr^2ω = (2/5)*0.4(*0.5^2)3 = 0.12 kgm^2/s The answer is:
A) 0.12 kgm^2/s
3. A flywheel (disk) with a mass of 8 kg and a radius of 0.3 m is spinning at an angular velocity of 4 rad/s. What is the angular momentum of the flywheel? (I = 1/2MR^2 for the flywheel)
A) 1.44 kgm^2/s B) 2.4 kgm^2/s C) 4.8 kgm^2/s D) 9.6 kgm^2/s
The solution is:
L = Iω = (1/2) mr^2ω = (1/2)*8(*0.3^2)4 = 1.44 kgm^2/s The answer is
A) 1.44 kgm^2/s
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Peter Apps
Kashvi Panjolia
Peter Apps
Kashvi Panjolia
Angular momentum is the rotational equivalent of linear momentum. It is represented by the equation L=I𝜔 where L is the angular momentum, I is the moment of inertia of the object, and 𝜔 is the angular velocity of the object. In addition, the angular momentum of an object moving in a straight line relative to a fixed point can be found by multiplying its linear momentum (p) by the perpendicular distance (r) to the point as shown in the picture.
🎥Watch: Minutephysics - What is Angular Momentum?
A student will slide a disk of mass:
The student wants the rod-disk system to end up with as much angular speed as possible.
CORRECT ANSWER: To the right of C
REASONING: To make the largest angular speed, the rod needs to get the greatest angular momentum possible (L=I𝜔). To get the largest angular momentum, the rod must be given the greatest possible torque. Since torque depends on the moment of inertia and the angular acceleration (𝜏=Iα), and the moment of inertia is calculated using I=cMR^2 (where c is a constant depending on the object), hitting the rod at the greatest radius will produce the greatest torque. Because the pivot is on the left side of C, we want to hit on the right side of C to maximize the radius so the moment of inertia increases, causing the net torque to increase, which will increase the angular speed.
ALTERNATE EXPLANATION: To create the largest angular speed, the rod needs to get the greatest angular momentum possible (L=I𝜔). To do this, we want the disk to transfer the greatest amount of momentum to the rod. Since angular momentum is conserved in the rod-disk system (no outside torques), this means that we want the initial angular momentum of the disk to be as large as possible. The angular momentum of the disk is equal to L=mvr where r is the distance from the pivot because the disk is moving in a straight-line motion relative to the pivot point of the rod.
To make:
(This explanation uses the law of conservation of momentum which will be covered in the next section.)
🎥Watch: AP Physics 1 - Unit 7 Streams
1. A spinning disk has a mass of 0.5 kg and a radius of 0.2 m. The disk is spinning at an angular velocity of 2 rad/s. What is the angular momentum of the disk? (I = 1/2MR^2 for the disk)
A) 0.02 kgm^2/s B) 0.2 kgm^2/s C) 0.4 kgm^2/s D) 0.8 kgm^2/s
The solution is: L = Iω = (1/2) mr^2ω = (1/2)*0.5(*0.2^2)2 = 0.02 kgm^2/s
The answer is A) 0.02 kgm^2/s
2. A 0.4 kg ball is attached to a string of length 0.5 m and is rotating in a circle with an angular velocity of 3 rad/s. What is the angular momentum of the ball? (I= 2/5MR^2 for the ball)
A) 0.3 kgm^2/s B) 0.12 kgm^2/s C) 0.9 kgm^2/s D) 1.2 kgm^2/s
The solution is:
L = Iω = (2/5) mr^2ω = (2/5)*0.4(*0.5^2)3 = 0.12 kgm^2/s The answer is:
A) 0.12 kgm^2/s
3. A flywheel (disk) with a mass of 8 kg and a radius of 0.3 m is spinning at an angular velocity of 4 rad/s. What is the angular momentum of the flywheel? (I = 1/2MR^2 for the flywheel)
A) 1.44 kgm^2/s B) 2.4 kgm^2/s C) 4.8 kgm^2/s D) 9.6 kgm^2/s
The solution is:
L = Iω = (1/2) mr^2ω = (1/2)*8(*0.3^2)4 = 1.44 kgm^2/s The answer is
A) 1.44 kgm^2/s
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