A small block on a frictionless surface has a mass

of 70 g. It is attached to a massless string passing
through a hole in a horizontal surface (see diagram).
The block is originally rotating in a circle of radius 45
cm with angular speed 0.80 rad/s. The string is then
pulled from below until the radius of the circle is 25
cm. You may treat the block as a point particle.

(a) Is the angular momentum of the block conserved?
Why or why not? (b) What is the final angular speed?
(c) What are the initial and final tensions in the string?
(d) What was the change in kinetic energy of the block?
(e) How much work was done in pulling the string?

Based on the fact that this is the exact same problem I am doing at the same time of the year I am guessing you might have been in physics 221 at Iowa state. If you are also in this class this may help you Paul, I have not figured out the tension part but

a) momentum is conserved
b) Paul already answered that one
c)?????
d)KE1=1/2*m(r1*w1)^2
KE2=1/2*m(r2*w2)^2
deltaKE=KE2-KE1
for this exact problem I got 101606.4 for the answer
e)work=change in KE

hopefully this helps others in this class of this and future semesters

I might be wrong but I may have just figured out how to find the tension even though it is not the way they wanted since the order is wrong but using the work we found we can probably use

W=Fd
and say the distance is the change in radius to get the force on the string. I hope this is the right answer and you don't have to calculate any forces caused by the blocks acceleration

just realized I forgot to turn 70g into kg so ignore the answer I gave for work It should be 101.61 (just moved the decimal over three places)

To answer these questions, we'll need to apply the principles of rotational motion and conservation of energy.

(a) Is the angular momentum of the block conserved?
Angular momentum is conserved if there is no external torque acting on the system. In this case, the only force that can exert a torque on the block is the tension in the string. Since the tension in the string changes when the radius decreases, there is an external torque acting on the block. Therefore, the angular momentum is not conserved.

(b) What is the final angular speed?
To find the final angular speed, we can use the principle of conservation of angular momentum. According to the principle, the initial angular momentum will be equal to the final angular momentum.

The initial angular momentum (L1) of the block is given by:
L1 = I1ω1

Where:
I1 = moment of inertia of the block about the axis of rotation
ω1 = initial angular speed

The final angular momentum (L2) of the block is given by:
L2 = I2ω2

Where:
I2 = moment of inertia of the block about the axis of rotation (changed due to reduced radius)
ω2 = final angular speed (to be calculated)

Since angular momentum is conserved, we have:
L1 = L2
I1ω1 = I2ω2

We know that the moment of inertia of a point particle rotating about an axis is given by:
I = mr^2
Where:
m = mass of the particle
r = radius

Using this equation, we can calculate the moment of inertia for both initial and final states. Given that the mass is 70g (0.07 kg) and the initial radius is 45cm (0.45 m), and the final radius is 25cm (0.25 m), we can calculate:

I1 = (0.07 kg)(0.45 m)^2
I2 = (0.07 kg)(0.25 m)^2

Now we can calculate the final angular speed (ω2) by rearranging the equation:
ω2 = (I1ω1) / I2

Substituting the values we calculated earlier, we can find the answer.

(c) What are the initial and final tensions in the string?
To determine the initial and final tensions in the string, we'll use the relationship between tension and centripetal force. The tension in the string provides the centripetal force to keep the block rotating in a circle.

The centripetal force (F) is given by:
F = mv^2/r

Where:
m = mass of the block
v = tangential velocity of the block
r = radius of the circle

For the initial state, the centripetal force is equal to the tension:
T1 = m(v1)^2/r1

For the final state, the centripetal force is equal to the tension:
T2 = m(v2)^2/r2

(v1) and (v2) can be calculated using the formula:
v = rω

Substituting the values of mass (m), initial and final radii (r1 and r2), and initial angular velocity (ω1), we can calculate the initial and final tensions (T1 and T2).

(d) What was the change in kinetic energy of the block?
The change in kinetic energy of the block is the difference between the initial and final kinetic energies.

The initial kinetic energy (K1) of the block is given by:
K1 = (1/2) I1 (ω1)^2

The final kinetic energy (K2) of the block is given by:
K2 = (1/2) I2 (ω2)^2

Substituting the values of the moments of inertia (I1 and I2) and initial and final angular speeds (ω1 and ω2), we can calculate the change in kinetic energy (ΔK).

ΔK = K2 - K1

(e) How much work was done in pulling the string?
To find the work done in pulling the string, we can use the work-energy theorem. According to this theorem, the work done on an object is equal to the change in its kinetic energy.

Since we already know the change in kinetic energy (ΔK) from part (d), the work done is simply:
Work = ΔK

Substitute the value of ΔK to calculate the amount of work done.

By following these steps and performing the calculations, you will be able to answer all the questions about the small block on a frictionless surface.