0.255 kg catcher is attached by a 45.0 cm long massless cable to a frictionless axle. The catcher is initially at rest. A 0.144 kg ball with a radius of 0.0875 m rolls without slipping from a height of 1.25 m and is captured by the catcher. For a solid ball, I = (2/5)MR2.
(a) What is the speed of the ball just before the ball is caught?
(b) What is linear speed of the ball and catcher immediately after the ball is caught? Only linear momentum is conserved in this collision
(c) If the cable is shortened to 21.5 cm, what would the linear speed of the ball and launcher be?
So for part a I got 4.18m/s and for part b I got 1.51m/s. For part c someone told me it would be the same as part b but when I asked my professor he said it would be different, and to use conservation momentum of angular speed, but I have no idea where to start. Any help would be appreciated!
To solve part (c) of the problem, we will need to consider the conservation of angular momentum.
When the cable is shortened to 21.5 cm, the moment of inertia of the entire system changes. To find the new angular speed, we need to observe the initial angular momentum and the final angular momentum of the system and equate them.
Let's define the initial situation as "i" and the final situation as "f."
In the initial situation (i), only the ball has rotational motion and the catcher is at rest. The angular momentum (L) of the system can be expressed as:
Li = Iball * ωball
where Iball is the moment of inertia of the ball and ωball is the angular speed of the ball.
In the final situation (f), both the ball and the catcher have linear motion. The catcher starts rotating about the axle with angular speed ω and the ball moves with linear speed v. The total angular momentum of the system can now be expressed as:
Lf = Iball * ωball + Icatcher * ωcatcher
Since the cable and axle are massless, we can assume that the moment of inertia of the catcher (Icatcher) is negligible compared to the ball. Therefore, the equation becomes:
Lf ≈ Iball * ωball
We know that angular momentum is conserved, so Li = Lf.
Therefore, we have:
Iball * ωball = Iball * ωball + Icatcher * ωcatcher
By rearranging the equation, we can find the final angular speed of the ball:
ωball = -Icatcher * ωcatcher / Iball
Now, we can use the relationship between linear speed (v) and the angular speed (ω) for an object undergoing pure rolling.
v = ω * r
where r is the radius of the ball.
So, the linear speed of the ball and the launcher can be expressed as:
vb = ωball * rb
vc = ωcatcher * rc
where rb is the radius of the ball and rc is the radius of the catcher.
Given the values provided in the question, you can substitute the appropriate values into the equations to find the linear speed of the ball and the launcher after shortening the cable to 21.5 cm.