Q. If the radius of a rotating disc is doubled while keeping the mass constant, how does the angular momentum change if the angular velocity remains the same?
A.It doubles
B.It remains the same
C.It quadruples
D.It halves
Solution
Angular momentum L = Iω; if radius is doubled, moment of inertia I increases by a factor of 4, hence L quadruples.
Q. Two particles A and B of masses m1 and m2 are moving in a circular path with angular velocities ω1 and ω2 respectively. What is the total angular momentum of the system?
A.m1ω1 + m2ω2
B.m1ω1 - m2ω2
C.m1ω1m2ω2
D.m1ω1 + m2ω2/2
Solution
Total angular momentum L = m1ω1 + m2ω2 for particles moving in the same direction.
Q. Two particles A and B of masses m1 and m2 are moving in a straight line with velocities v1 and v2 respectively. If they collide elastically, which of the following statements is true regarding their angular momentum about the center of mass?
A.It is conserved
B.It is not conserved
C.Depends on the masses
D.Depends on the velocities
Solution
Angular momentum about the center of mass is conserved in an elastic collision.
Q. Two particles A and B of masses m1 and m2 are moving in opposite directions with velocities v1 and v2 respectively. What is the total angular momentum of the system about the origin if they are at a distance r from the origin?
A.m1v1r + m2v2r
B.m1v1r - m2v2r
C.m1v1r + m2(-v2)r
D.0
Solution
Total angular momentum L = m1v1r - m2v2r, but since they are in opposite directions, it simplifies to m1v1r + m2v2r.
Q. Two particles A and B of masses m1 and m2 are moving in opposite directions with velocities v1 and v2 respectively. What is the total angular momentum of the system about a point O located at the midpoint between A and B?
A.(m1v1 + m2v2)r
B.(m1v1 - m2v2)r
C.0
D.(m1v1 + m2v2)/2
Solution
Since they are moving in opposite directions, the total angular momentum about point O is zero.
Q. Two particles A and B of masses m1 and m2 are moving in opposite directions with velocities v1 and v2 respectively. What is the total angular momentum of the system about the origin?
A.m1v1 + m2v2
B.m1v1 - m2v2
C.m1v1 + m2(-v2)
D.m1v1 + m2v2
Solution
Total angular momentum L = m1v1 + m2(-v2) = m1v1 - m2v2.
Q. Two particles A and B of masses m1 and m2 are moving in opposite directions with velocities v1 and v2 respectively. What is the total angular momentum of the system about a point O located at the center of mass?
A.(m1v1 + m2v2)
B.(m1v1 - m2v2)
C.m1v1 + m2v2
D.0
Solution
Total angular momentum is the sum of individual angular momenta, which is m1v1 + m2v2.
Q. Two particles A and B of masses m1 and m2 are moving with velocities v1 and v2 respectively. If they collide elastically, which of the following statements is true regarding their angular momentum about the center of mass?
A.It is conserved
B.It is not conserved
C.Depends on the masses
D.Depends on the velocities
Solution
Angular momentum is conserved in an elastic collision about the center of mass.
Q. Two particles of masses m1 and m2 are moving in a circular path of radius r with angular velocities ω1 and ω2 respectively. What is the total angular momentum of the system?
Q. Two particles of masses m1 and m2 are moving in a circular path with radii r1 and r2 respectively. If they have the same angular velocity, what is the ratio of their angular momenta?
A.m1r1/m2r2
B.m1/m2
C.r1/r2
D.m1r2/m2r1
Solution
Angular momentum L = mvr, thus L1/L2 = (m1r1)/(m2r2) when ω is constant.
Q. Two particles of masses m1 and m2 are moving in a straight line with velocities v1 and v2 respectively. If they collide elastically, what is the expression for the change in angular momentum about the center of mass?
A.m1v1 + m2v2
B.m1v1 - m2v2
C.0
D.m1v1 + m2v2 - (m1v1' + m2v2')
Solution
In an elastic collision, the total angular momentum about the center of mass is conserved.
