Angular momentum and change in velocity

[physgrl]

Homework Statement

A turntable has a moment of inertia of 2.5 x 10-2 kg m2 and spins freely at 33.3 rev/min. A 0.25 kg ball of putty is dropped vertically on the turntable at a point 0.20 m from the center. What is the new angular speed of the system?

a. 41 rev/min
b. 24 rev/min
c. 33 rev/min
d. 27 rev/min

Homework Equations

L=Iω
p=mv
L=r*p

The Attempt at a Solution

i tried to do Iω=(I+mr)ω(final)
i got 27 rev/min
the real answer is supposed to be 24rev/min according to the answer key

 

[hage567]

Homework Statement

A turntable has a moment of inertia of 2.5 x 10-2 kg m2 and spins freely at 33.3 rev/min. A 0.25 kg ball of putty is dropped vertically on the turntable at a point 0.20 m from the center. What is the new angular speed of the system?

a. 41 rev/min
b. 24 rev/min
c. 33 rev/min
d. 27 rev/min

Homework Equations

L=Iω
p=mv
L=r*p

The Attempt at a Solution

i tried to do Iω=(I+mr)ω(final)
i got 27 rev/min
the real answer is supposed to be 24rev/min according to the answer key

Can you show the details of your calculation? We can't tell you what you did wrong if you don't show us your work!

 

[LawrenceC]

"i tried to do Iω=(I+mr)ω(final)"

Careful, mr does not have same units as I.

 

[physgrl]

Yeah that was my error thanks!

 

[asteriatic]

Based on the given information, the initial angular momentum of the system can be calculated using the equation L=Iω, where I is the moment of inertia and ω is the angular velocity. Substituting the values, we get L= 2.5 x 10-2 kg m2 * (33.3 rev/min * 2π rad/rev) = 5.24 x 10-2 kg m2/s.

When the 0.25 kg ball of putty is dropped onto the turntable, the system experiences a change in angular momentum. This can be calculated using the equation ΔL= r * Δp, where r is the distance from the center of the turntable to the point of impact and Δp is the change in linear momentum of the ball. Since the ball is dropped vertically, there is no change in horizontal momentum and Δp= mv= 0.25 kg * 9.8 m/s2 * 0.2 m = 0.49 kg m/s.

Substituting the values in the equation, we get ΔL= 0.2 m * 0.49 kg m/s = 0.098 kg m2/s.

According to the law of conservation of angular momentum, the total angular momentum of a system remains constant unless an external torque is applied. Therefore, the initial angular momentum of the system must be equal to the final angular momentum after the ball is dropped.

We can use the equation Iω=(I+mr)*ω(final) to calculate the final angular velocity of the system. Substituting the values, we get (2.5 x 10-2 kg m2 * 33.3 rev/min * 2π rad/rev) = (2.5 x 10-2 kg m2 + 0.25 kg * 0.2 m) * ω(final). Solving for ω(final), we get ω(final)= 24.95 rev/min.

Therefore, the new angular speed of the system after the ball is dropped is approximately 24 rev/min, which matches with option b in the given choices. This confirms that the answer key is correct.