Question

OUESTION1:

A 250-kg beam 2.6 m in length slides broadside down the ice with a speed of 22 m/s. A 63-kg man at rest grabs one end as it goes past and hangs on as both he and the beam go spinning down the ice. Assume frictionless motion.

a-)How fast does the center of mass of the system move after the collision?

b-)With what angular velocity does the system rotate about its cm?

QUESTION-02

The uniform bar shown in has a length of 0.80 m. The bar begins to rotate from rest in the horizontal plane about the axis passing through its left end. What will be the magnitude of the angular momentum L of the bar 6.0 s after the motion has begun?

OUESTION-03

Consider a turntable to be a circular disk of moment of inertia It rotating at a constant angular velocity ωi around an axis through the center and perpendicular to the plane of the disk (the disk's "primary axis of symmetry"). The axis of the disk is vertical and the disk is supported by frictionless bearings. The motor of the turntable is off, so there is no external torque being applied to the axis.
Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia of the record is Ir. The initial angular velocity of the second disk is zero.

There is friction between the two disks.

After this "rotational collision," the disks will eventually rotate with the same angular velocity.

Because of friction, rotational kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final rotational kinetic energy, Kf, of the two spinning disks?

Express the final kinetic energy in terms of It, Ir, and the initial kinetic energy Ki of the two-disk system. No angular velocities should appear in your answer.

Answer 1

1) (a) Im assuming "collision", based on the information you have given, is the man grabbing the bar.
momentum of system (assumption linear momentum) => initial momentum = final momentum if conservation of momentum is applied.
therefore p = mv
p initial = 250 x 22 = 5500 kgms^-2

p final = (250 + 63) x vfinal = p intial
5500 = 313*v
v = 17.6 m/s

(b) Equation: ω=v/r (ω=angular velocity)
as the beam is 2.6 metres, the centre of mass of the beam is 1.3 metre (therefore the radius is 1.3 metre) therefore:
ω= 17.6/1.3 = 13.5 rad/s

HOWEVER im not sure how your teacher wants the answer because the centre of mass of the SYSTEM changes when that man hangs onto the end. its adding 63 kg to the system therefore the centre of mass changes. the NEW radius is 1.56m therefore the angular velocity should be

17.5/1.56 = 11.22 rad/s

to get 1.56 m as the new center of mass, which will be the pivot at which the beam will spin, you use moments to place the centre of mass.

draw a straight line, label it 2.6 metres. at 1.3 metre (halfway) draw an arrow facing down and write 250 kg. on one of the extreme ends of the beam draw an arrow pointing down and write 63 kg. now the centre of mass is the point at which the average mass of the whole system is, so a fulcrum needs to be placed such that the system balances... so draw a triangle below the line between 250 and 63 arrows, draw it closer to 250 than 63. label x the distance from the triangle (fulcrum) to 63 and the distance from 250 to the triangle will be (1 - x). this is because the distance from the 250 point to 63 kg arrow is 1.3 metre. now use moments, momentum = Fd

(250x9.81) x (1.3-x) = (63x9.81) x (x)
they equate because the sum of the anti clockwise and clockwise moments will be zero if it is balanced. (which is where the centre of mass will be)
so out of that you will get x (1.04) and 1.3-x (0.26). you use the 0.26 distance to add onto the 1.3 metre as it will form a bigger circle when spun from that point.

2) The angular acceleration α = T/I, where T is the total torque and I is the moment of inertia. The total torque is 0.8*8 - 0.6*12 N*m.
α = (0.8*8 - 0.6*12) / I
The angular velocity after time t is ω = α*t, and the angular momentum L = ω*I = T/I *t * I = T * t
L = (0.8*8 - 0.6*12)*(-6) = 4.8