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.