A thin uniform bar, hanging, is hit by a ball. Find the angular velocity

[Phys121VIU]

Homework Statement

A thin, uniform metal bar, 3m long and weighing 75N, is hanging veritcally from the ceiling by a frictionless pivot. Suddenly it is struck 1.3m below the ceiling by a small 2 kg clay ball, initially traveling horizontally at 14m/s. The ball sticks to the bar

Find the angular velocity of the ball+bar just after the collision.

Homework Equations

The Attempt at a Solution

Angular momentum is conserved so my understanding is that

Angular momentum of the ball = angular momentum of the bar and bar

L_ball = (I_bar + I_ball)ω²

(2kg)(14m/s)(1.3m) = [(1/3)(75N/9.8m/s²)(3m)² + (2kg)(1.3m)²]ω²

so i worked it out and got ω = 1.18 rad/s

but that's not correct...I can't think of any other way to approach it

 

[ehild]

Homework Statement

A thin, uniform metal bar, 3m long and weighing 75N, is hanging veritcally from the ceiling by a frictionless pivot. Suddenly it is struck 1.3m below the ceiling by a small 2 kg clay ball, initially traveling horizontally at 14m/s. The ball sticks to the bar

Find the angular velocity of the ball+bar just after the collision.

Homework Equations

The Attempt at a Solution

Angular momentum is conserved so my understanding is that

Angular momentum of the ball = angular momentum of the bar and bar

L_ball = (I_bar + I_ball)ω²

Are you sure that the last formula is correct?

ehild

 

[Phys121VIU]

Wow..its definitely not squared..thanks:P

Found the answer..1.38rad/s!

 

[ehild]

Well done !

 

[Nickie]

though.

Your approach is correct, however there are a few errors in your calculation. The moment of inertia of a thin uniform bar is (1/12)ML^2, not (1/3)ML^2. Also, the units for moment of inertia should be in kgm^2, not Nm^2. Additionally, the moment of inertia for a point mass (the clay ball) is simply MR^2, where R is the distance from the pivot point. So the correct equation should be:

(2kg)(14m/s)(1.3m) = [(1/12)(75kg)(3m)^2 + (2kg)(1.3m)^2]ω

Solving for ω, we get ω = 2.23 rad/s. This is the angular velocity of the ball+bar system just after the collision. Keep in mind that the ball+bar system will continue to rotate at this angular velocity until acted upon by an external force.