Suppose three masses are arranged as shown, connected by a rodless mass.


(2 kg)-------(6 kg)--------(4 kg)

2m 3m

a) If this object is free to rotate in space, about what point will it spin?

b) A small mass (m=0.1 kg) drops vertically onto the2 kg mass traveling 100 m/s and buries intself into the mass. What is the period of rotation of the system immediately after impact?

c) What is the rotational kinectic energy of the system after the impact?

I'm sorry, 2m is the distance betweek 2 kg and 6 kg. 3m is the distance between 6kg and 4kg!

a. it spins about its center of mass. Where is that? From the left end,

cm=(2*0+6*2+4*5)/12kg=28/12 m from the left end.

b. conservation of angular momentum:
momentum before=momentum after
Io (v/r)=Itotal*w
.1*r^2*100/r=Itotal w
now what is r? well, r is the distance from the left end to the cm, or 28/12 m
What is I total?
it is the sum of the three masses.
I total= Ileft + I big + I right
= (2.1kg)(28/12)^2+ 6kg*(28/12-2)^2+ 4kg*(5-28/12)^2

solve for w, in radians per second.

Period= 2PI/w

a) To determine about which point the object will spin, we need to find the center of mass of the system. The center of mass is the point where the average position of all the mass is located. It can be found by considering the individual masses and their distances from a reference point.

In this case, let's take the reference point as the leftmost position of the system. The center of mass (CM) can be calculated using the formula:

CM = (m1 * r1 + m2 * r2 + m3 * r3) / (m1 + m2 + m3)

where m1, m2, and m3 are the masses, and r1, r2, and r3 are their respective distances from the reference point.

For the given system:
m1 = 2 kg, r1 = 2m
m2 = 6 kg, r2 = 5m (distance from the 2kg mass)
m3 = 4 kg, r3 = 8m (distance from the 6kg mass)

Plugging in the values, the center of mass can be calculated as:

CM = (2 kg * 2m + 6 kg * 5m + 4 kg * 8m) / (2 kg + 6 kg + 4 kg)

Simplifying the equation:

CM = (4 kg + 30 kg + 32 kg) / 12 kg

CM = 66 kgm / 12 kg

CM = 5.5 m

Therefore, the object will spin about the point located 5.5 meters to the right of the leftmost position.

b) To calculate the period of rotation of the system immediately after impact, we need to use the principle of conservation of angular momentum. Angular momentum is conserved if no external torques are acting on the system.

Initially, the system has no angular momentum since it is at rest. When the small mass (0.1 kg) drops vertically onto the 2 kg mass, it transfers its linear momentum to the system, and the system starts rotating.

The formula for angular momentum is given by:

angular momentum (L) = moment of inertia (I) * angular velocity (ω)

The moment of inertia (I) for a system of multiple masses connected by a rodless mass is given by:

I = Σ(mass * distance^2)

The initial angular momentum of the system after the impact is equal to the angular momentum of the dropped mass, which can be calculated as:

L_0 = I_dropped_mass * ω_0

The angular velocity (ω_0) can be calculated using the relation between linear velocity (v) and angular velocity (ω):

v = r * ω

Initially, the dropped mass was traveling at 100 m/s and gets buried into the 2 kg mass. To find the angular velocity, we need to determine the new linear velocity of the combined system.

Using the conservation of linear momentum:

(mass_dropped_mass * initial_velocity_dropped_mass) = (mass_combined_system * final_velocity_combined_system)

(0.1 kg * 100 m/s) = ((2 kg + 0.1 kg) * final_velocity_combined_system)

10 kg⋅m/s = (2.1 kg) * final_velocity_combined_system

final_velocity_combined_system = 10 kg⋅m/s / 2.1 kg

final_velocity_combined_system = 4.76 m/s (approx)

Therefore, the linear velocity of the combined system after the impact is 4.76 m/s.

Now, we can calculate the initial angular momentum (L_0) using the moment of inertia (I) and the initial angular velocity (ω_0):

I = (2 kg * 2m^2) + (6 kg * 5m^2) + (4 kg * 8m^2)

I = 4 kgm^2 + 150 kgm^2 + 256 kgm^2

I = 410 kgm^2

ω_0 = v / r

ω_0 = 4.76 m/s / 5m

ω_0 = 0.952 rad/s

L_0 = 410 kgm^2 * 0.952 rad/s

L_0 = 391.12 kgm^2/s

Now, to calculate the period (T), we can use the formula:

T = 2π / ω

T = 2π / 0.952 rad/s

Calculating the period:

T ≈ 6.60 seconds (approx)

Therefore, the period of rotation of the system immediately after impact is approximately 6.60 seconds.

c) The rotational kinetic energy of the system after the impact can be calculated using the formula:

Rotational kinetic energy (K) = 0.5 * moment of inertia (I) * angular velocity squared (ω^2)

The moment of inertia (I) for the combined system can be found using the formula mentioned in part (b).

I = (2 kg * 2m^2) + (6 kg * 5m^2) + (4 kg * 8m^2)

I = 4 kgm^2 + 150 kgm^2 + 256 kgm^2

I = 410 kgm^2

Using the angular velocity (ω) obtained in part (b):

ω = 0.952 rad/s

Calculating the rotational kinetic energy:

K = 0.5 * 410 kgm^2 * (0.952 rad/s)^2

K ≈ 185.07 J (approx)

Therefore, the rotational kinetic energy of the system after the impact is approximately 185.07 Joules.