Q. A rigid body is rotating about a fixed axis with an angular velocity ω. If the moment of inertia of the body is I, what is the angular momentum of the body?
A.Iω
B.ω/I
C.I/ω
D.Iω^2
Solution
Angular momentum L = Iω, where I is the moment of inertia and ω is the angular velocity.
Q. A rigid body is rotating about a fixed axis. If the moment of inertia of the body is I and it is rotating with an angular velocity ω, what is its angular momentum?
A.Iω
B.I/ω
C.Iω^2
D.ω/I
Solution
Angular momentum L = Iω, where I is the moment of inertia and ω is the angular velocity.
Q. A rigid body rotates about a fixed axis with an angular velocity ω. If the moment of inertia of the body is I, what is the angular momentum of the body?
A.Iω
B.ω/I
C.I/ω
D.Iω^2
Solution
Angular momentum L = Iω, where I is the moment of inertia and ω is the angular velocity.
Q. A roller coaster at the top of a hill has a potential energy of 5000 J. If it descends to a height of 10 m, what is its speed at the bottom? (g = 9.8 m/s²)
A.10 m/s
B.20 m/s
C.30 m/s
D.40 m/s
Solution
Using conservation of energy, initial PE + KE = final PE + KE. 5000 J = mgh + 0.5mv². Solving gives v = √(2(5000 - mgh)/m) = 30 m/s.
Q. A roller coaster starts from rest at a height of 30 m. What is its speed at the lowest point? (g = 9.8 m/s²)
A.10 m/s
B.15 m/s
C.20 m/s
D.25 m/s
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 30) = 24.5 m/s.
Q. A roller coaster starts from rest at a height of 50 m. What is its speed at the lowest point?
A.10 m/s
B.20 m/s
C.30 m/s
D.40 m/s
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*50) = 31.3 m/s.
Q. A rolling object has both translational and rotational motion. Which of the following quantities remains constant for a rolling object on a flat surface?
A.Linear velocity
B.Angular velocity
C.Total energy
D.Kinetic energy
Solution
The total energy remains constant for a rolling object on a flat surface, assuming no external work is done.
Q. A rotating disc has an angular velocity of ω. If the radius of the disc is doubled while keeping the mass constant, what happens to the angular momentum?
A.It doubles
B.It remains the same
C.It quadruples
D.It halves
Solution
Angular momentum L = Iω, where I is the moment of inertia. If radius is doubled, I increases by a factor of 4, but ω decreases by a factor of 2, so L remains the same.
Q. A rotating object has an angular momentum L. If the moment of inertia of the object is doubled while keeping the angular velocity constant, what happens to the angular momentum?
A.It doubles
B.It halves
C.It remains the same
D.It quadruples
Solution
Angular momentum L = Iω. If I is doubled and ω remains constant, L also doubles.
Q. A rotating object has an angular momentum of L. If its angular velocity is doubled and its moment of inertia remains constant, what will be the new angular momentum?
A.L
B.2L
C.4L
D.L/2
Solution
Angular momentum L = Iω, if ω is doubled, L becomes 2I(2ω) = 4L.
Q. A rotating object has an angular momentum of L. If its moment of inertia is doubled while keeping the angular velocity constant, what will happen to its angular momentum?
A.It doubles
B.It halves
C.It remains the same
D.It becomes zero
Solution
Angular momentum L = Iω; if I is doubled and ω remains constant, L remains the same.
Q. A rotating object has an angular momentum of L. If its moment of inertia is halved and its angular velocity is doubled, what is the new angular momentum?
Q. A rotating object has an angular momentum of L. If its moment of inertia is halved and the angular velocity is doubled, what is the new angular momentum?
A.L
B.2L
C.4L
D.L/2
Solution
New angular momentum L' = I'ω' = (1/2 I)(2ω) = Iω = L.
