A small block on a frictionless horizontal surface has a mass of 2.50×10−2 Kg. It is attached to a massless cord passing through a hole in the surface. The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 1.75 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. You may treat the block as a particle.
A) is angular momentum conserved?
B) What is the new angular speed?
C) Find the change in kinetic energy of the block.
D) how much work was done in pulling the cord.
For C i have:
KEf=(1/2)(2.5*10^-2)(.150*7)^2
=.01378125
KEi=(1/2)(2.5*10^-2)(.300*1.75)^2
=.0034453125
That's correct. Try to think about why angular momentum is conserved and why kinetic energy is not conserved.
You can also consider a similar problem where the cord is not pulled but instead it is wrapping itself around a stick. What happens to angular momentum and kinetic energy in this case?
how do you find B and D then?
B: form conseravtion of momentum, which yields the angular velocities of 7 rad/sec that you used.
D follows from conservation of energy. The work done equals the increase in kinetic energy.
Typo: momentum --> angular momentum
A) No, angular momentum is not conserved in this situation. When the cord is pulled, the radius of the circle decreases, which means the moment of inertia of the block changes. Since angular momentum is the product of moment of inertia and angular speed, any change in moment of inertia or radius will cause a change in angular momentum.
B) To find the new angular speed, we can use the principle of conservation of angular momentum. The initial angular momentum is given by the product of the moment of inertia and initial angular speed, and the final angular momentum is given by the product of the new moment of inertia and the new angular speed. Since the moment of inertia changes, we can set up the equation:
(Ii * ωi) = (If * ωf)
where Ii is the initial moment of inertia, ωi is the initial angular speed, If is the final moment of inertia, and ωf is the final angular speed.
C) To find the change in kinetic energy, we need to calculate the initial and final kinetic energies and then find the difference.
Initial kinetic energy (KEi) can be calculated using the formula:
KEi = (1/2) * mass * (initial velocity)^2
Final kinetic energy (KEf) can be calculated using the formula:
KEf = (1/2) * mass * (final velocity)^2
The change in kinetic energy (ΔKE) can be calculated as:
ΔKE = KEf - KEi
D) To find the work done in pulling the cord, we can use the formula:
Work = force * displacement * cos(θ)
where force is the tension in the cord, displacement is the distance the cord was pulled, and θ is the angle between the direction of force and displacement.
To answer the given questions:
A) To determine if angular momentum is conserved, we need to check if the angular momentum before and after the cord is pulled remains the same. The angular momentum (L) of an object can be calculated using the formula L = Iω, where I is the moment of inertia and ω is the angular speed.
Before the cord is pulled, the moment of inertia (I) can be calculated using the formula I = mr^2, where m is the mass of the block and r is the initial radius.
I = (2.50×10^(-2) kg) * (0.300 m)^2 = 2.25×10^(-3) kg * m^2
The initial angular momentum (Li) is then Li = I * ω = (2.25×10^(-3) kg * m^2) * (1.75 rad/s) = 3.9375×10^(-3) kg * m^2/s
After the cord is pulled, the block rotates with a new radius of 0.150 m. The final moment of inertia (If) can be calculated the same way:
If = (2.50×10^(-2) kg) * (0.150 m)^2 = 5.625×10^(-4) kg * m^2
The final angular momentum (Lf) is then Lf = If * ωf, where ωf is the new angular speed. Since we don't know ωf yet, we can't determine if angular momentum is conserved at this point.
B) To find the new angular speed (ωf), we can use the principle of conservation of angular momentum. According to this principle, angular momentum is conserved if no external torque acts on the system. In this case, since there is no mention of any external torque, we assume angular momentum is conserved.
Since Li = Lf, we can equate the expressions:
Li = Lf
I * ω = If * ωf
(2.25×10^(-3) kg * m^2) * (1.75 rad/s) = (5.625×10^(-4) kg * m^2) * ωf
Solving for ωf gives us:
ωf = (2.25×10^(-3) kg * m^2 * 1.75 rad/s) / (5.625×10^(-4) kg * m^2) ≈ 8.33 rad/s
Thus, the new angular speed is approximately 8.33 rad/s.
C) To find the change in kinetic energy of the block, we can calculate the initial and final kinetic energies using the formulas:
Initial Kinetic Energy (KEi) = (1/2) * m * v^2
Final Kinetic Energy (KEf) = (1/2) * m * v^2
The block is revolving in a circle, so its velocity (v) can be calculated as the product of the angular speed (ω) and the radius (r).
KEi = (1/2) * (2.50×10^(-2) kg) * ((0.300 m) * (1.75 rad/s))^2 = 0.00344846 Joules
KEf = (1/2) * (2.50×10^(-2) kg) * ((0.150 m) * (8.33 rad/s))^2 = 0.00545312 Joules
The change in kinetic energy (ΔKE) can be calculated by subtracting the initial kinetic energy from the final kinetic energy:
ΔKE = KEf - KEi = 0.00545312 Joules - 0.00344846 Joules = 0.00200466 Joules
Therefore, the change in kinetic energy of the block is approximately 0.002005 Joules.
D) The work done in pulling the cord can be calculated using the work-energy theorem. The work done (W) is equal to the change in kinetic energy (ΔKE). Hence, the work done in pulling the cord is the same as the change in kinetic energy, which is approximately 0.002005 Joules.