A uniform rod of mass M and length d is initially at rest on a horizontal and frictionless table contained in the xy plane, the plane of the screen. The figure is a top view, gravity points into the screen. The rod is free to rotate about an axis perpendicular to the plane and passing through the pivot point at a distance d/3 measured from one of its ends as shown. A small point mass m, moving with speed v0, hits the rod and stick to it at the point of impact at a distance d/3 from the pivot.

(a) If the mass of the rod is M=4m, what is the magnitude of the angular velocity of the rod+small mass system after the collision?

1)ω=2v0/(3d)
2)ω=3v0/(2d)
3)ω=4v0/(3d)
4)ω=3v0/(4d)
5)ω=8v0/(3d)
6)ω=3v0/(8d)
7)ω=5v0/(3d)
8)ω=3v0/(5d)

Using again M=4m. What is the speed of the center of mass of the rod right after collision?

1)vcm=v0
2)vcm=v0/2
3)vcm=v0/3
4)vcm=v0/5
5)vcm=v0/10
6)vcm=v0/20

distance of Center of rod from pivot = 3d/6 - 2d/6 = d/6

so
I about pivot = (1/12)M d^2 + M d^2/36
= (1/9)Md^2 = (4/9)m d^2

distance of mass trajectory from pivot = d/3

angular momentum of m about pivot before collision = m vo r = m vo d/3
= total ang momentum of system about pivot before and after collision

angular momentum of m about pivot after collision = m (d/3)^2 w = m d^2 w/9

angular momentum of rod about pivot after collision = [(1/12)M d^2 + M d^2/36] w
=(Md^2 /9)w
so
m vo d/3 = M d^2 w/9 + m d^2 w/9
if M = 4 m

m vo /3 = 4 m d w/9 + m d w/9

vo = 4 d w/3 + d w/3 = 5 d w/3

w = 3 vo / 5d

w r = (3 v0/5d)(d/6) = vo/10

Thank you for giving the steps, I was lost in this question.

A mokey of mass 15 kg is climbing on a rope with an acceleration of 1 m/s square, how much force should it apply to the rope ? If the rope is 5 m long an the monkey starts from rest, how time will it take to reach the ceiling?

To solve this problem, we need to apply the principle of conservation of angular momentum and the principle of conservation of linear momentum.

(a) Let's first calculate the angular velocity of the rod+small mass system after the collision.

The principle of conservation of angular momentum states that the total angular momentum before the collision is equal to the total angular momentum after the collision.

Initially, the rod is at rest, so its initial angular momentum is zero.

The point mass m hits the rod and sticks to it at a distance d/3 from the pivot. The total moment of inertia of the system will be the sum of the moment of inertia of the rod and the moment of inertia of the point mass attached to it.

The moment of inertia of the rod rotating about an axis perpendicular to the xy plane and passing through the pivot point at a distance d/3 from one of its ends is given by:

I_rod = (1/3) * M * (d/3)^2
= M * d^2 / 27

The moment of inertia of the point mass attached to the rod, considering it rotates about the same axis, is given by:

I_pointmass = m * (d/3)^2
= m * d^2 / 9

The total moment of inertia of the system is the sum of these two:

I_total = I_rod + I_pointmass
= M * d^2 / 27 + m * d^2 / 9
= (4m * d^2 / 27) + (m * d^2 / 9)
= (4md^2 + 9md^2) / 27
= 13md^2 / 27

Now, using the principle of conservation of angular momentum, we equate the initial angular momentum to the final angular momentum:

0 = (I_rod + I_pointmass) * ω_final

Solving for ω_final:

ω_final = 0 / (I_rod + I_pointmass)
= 0

Therefore, the magnitude of the angular velocity of the rod+small mass system after the collision is ω = 0 rad/s.

Answer: (a) ω = 0 rad/s.

Using the same value of M = 4m, let's now calculate the speed of the center of mass of the rod right after the collision.

The principle of conservation of linear momentum states that the total linear momentum before the collision is equal to the total linear momentum after the collision.

Initially, the system is stationary, so the initial linear momentum is zero.

The point mass m hits the rod and sticks to it, transferring its linear momentum to the rod+small mass system. Let's denote the speed of the point mass m as v_m.

The total mass of the system is M+ m.

Using the principle of conservation of linear momentum, we equate the initial linear momentum to the final linear momentum:

0 = (M + m) * 0 + m * v_0

Solving for v_0:

v_0 = 0 / (M + m)
= 0

Therefore, the speed of the center of mass of the rod right after the collision is v_cm = 0 m/s.

Answer: (b) v_cm = 0 m/s.