A skater spinning with angular speed of 1.5 rad/s draws in her outstretched arms thereby reducing her moment of inertia by a factor a 3.
a) Find her final angular speed.
b)What is the ratio of her final angular momentum to her initial angular momentum?
c) Determine the ratio of her final kinetic energy to her initial kinetic energy.
d)Did her mechanical energy change? (Yes but why?)
This is very similar to another question I answered earlier today. Use the same approach.
http://www.jiskha.com/display.cgi?id=1291313864
Thank You!
To solve this problem, we'll use the conservation of angular momentum and the conservation of energy.
a) The initial angular speed is 1.5 rad/s, and the moment of inertia is reduced by a factor of 3. Let's call the final angular speed "ωf." According to the conservation of angular momentum, the initial angular momentum (L_initial) is equal to the final angular momentum (L_final).
L_initial = L_final
I_initial * ω_initial = I_final * ω_final
Since the moment of inertia is reduced by a factor of 3, we can write:
(1/3 * I_initial) * ω_initial = I_final * ω_final
Now, substitute the given values:
(1/3) * (I_initial) * 1.5 rad/s = I_final * ω_final
Simplifying, we find:
0.5 * ω_initial = ω_final
Therefore, the final angular speed is 0.5 rad/s.
b) The ratio of the final angular momentum (L_final) to the initial angular momentum (L_initial) can be calculated as:
(L_final / L_initial) = (I_final * ω_final) / (I_initial * ω_initial)
Substituting the given values:
(L_final / L_initial) = (I_final * 0.5 rad/s) / (I_initial * 1.5 rad/s)
Since L_final and L_initial are the same, this reduces to:
1 = (I_final * 0.5) / (I_initial * 1.5)
Multiplying both sides by (I_initial * 1.5), we get:
I_initial * 1.5 = I_final * 0.5
Dividing both sides by (I_initial * 0.5), we find:
3 = I_final / I_initial
Therefore, the ratio of the final angular momentum to the initial angular momentum is 3.
c) The ratio of the final kinetic energy (K_final) to the initial kinetic energy (K_initial) can be calculated as:
(K_final / K_initial) = (1/2 * I_final * ω_final^2) / (1/2 * I_initial * ω_initial^2)
Simplifying, we find:
(K_final / K_initial) = (I_final * ω_final^2) / (I_initial * ω_initial^2)
Substituting the given values:
(K_final / K_initial) = (I_final * (0.5 rad/s)^2) / (I_initial * (1.5 rad/s)^2)
(K_final / K_initial) = (I_final * 0.25) / (I_initial * 2.25)
(K_final / K_initial) = I_final / (I_initial * 9)
Since we know the ratio of I_final to I_initial is 1/3, this becomes:
(K_final / K_initial) = (1/3) / (I_initial * 9)
Therefore, the ratio of the final kinetic energy to the initial kinetic energy is 1/27.
d) Yes, the mechanical energy changed. The mechanical energy of a spinning skater consists of both potential energy (due to the distance of mass from the axis of rotation) and kinetic energy (due to the motion of the skater). Reducing the moment of inertia by drawing in her arms decreases the potential energy, but the kinetic energy remains constant. Thus, the mechanical energy decreases.
a) To find the final angular speed of the skater, we can use the law of conservation of angular momentum. According to this law, the initial angular momentum of a system remains constant as long as there are no external torques acting on it. The formula for angular momentum is given by:
L = Iω,
where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
In this case, the skater reduces her moment of inertia by a factor of 3. Therefore, her new moment of inertia (I') becomes 1/3 of the initial moment of inertia (I). Mathematically, we can express this as I' = (1/3)I.
Since the angular momentum is conserved, we can set the initial angular momentum equal to the final angular momentum:
LI = L'I',
where L is the initial angular momentum and L' is the final angular momentum.
Substituting the expressions for the angular momenta and the moment of inertia, we get:
Iω = I'(ω'),
where ω' is the final angular speed we want to find.
Plugging in the given values (ω = 1.5 rad/s and I' = (1/3)I), we can solve for ω':
I(1.5) = (1/3)I(ω')
Simplifying the equation gives:
1.5 = (1/3)ω'
Multiply both sides by 3:
4.5 = ω'
Therefore, the skater's final angular speed is 4.5 rad/s.
b) To find the ratio of the final angular momentum to the initial angular momentum, we can use the formula for angular momentum (L = Iω) that we used in part a).
The ratio can be expressed as:
(L')/(L) = (I'ω')/(Iω),
where L is the initial angular momentum, L' is the final angular momentum, I is the initial moment of inertia, I' is the final moment of inertia, ω is the initial angular speed, and ω' is the final angular speed.
We already know the values of ω and ω' from part a), and we know that I' = (1/3)I. Therefore, substituting these values into the expression, we get:
(L')/(L) = ((1/3)Iω')/(Iω),
Canceling out the common terms, we get:
(L')/(L) = (1/3)(ω')/(ω),
Substituting the values we found in part a), we can solve for the ratio:
(L')/(L) = (1/3)(4.5)/(1.5) = 1.
Therefore, the ratio of the skater's final angular momentum to her initial angular momentum is 1.
c) The ratio of her final kinetic energy to her initial kinetic energy can be found using the formula for kinetic energy (K = (1/2)Iω^2).
The expression for the ratio can be written as:
(K')/(K) = ((1/2)I'ω'^2)/((1/2)Iω^2),
where K is the initial kinetic energy, K' is the final kinetic energy, I is the initial moment of inertia, I' is the final moment of inertia, ω is the initial angular speed, and ω' is the final angular speed.
From part a), we know the values of ω and ω', and we know that I' = (1/3)I. Substituting these values into the expression, we get:
(K')/(K) = ((1/2)(1/3)Iω'^2)/((1/2)Iω^2),
Canceling out the common terms, we get:
(K')/(K) = (1/3)(ω'^2)/(ω^2).
Using the values we found in part a), we can solve for the ratio as:
(K')/(K) = (1/3)(4.5^2)/(1.5^2) = 3.
Therefore, the ratio of the skater's final kinetic energy to her initial kinetic energy is 3.
d) Yes, her mechanical energy changed. Mechanical energy is the sum of kinetic energy and potential energy. In this case, we are only considering kinetic energy.
From part c), we found that the ratio of her final kinetic energy to her initial kinetic energy is 3. Since the ratio is greater than 1, it means that her final kinetic energy is greater than her initial kinetic energy.
Therefore, her mechanical energy changed because her kinetic energy increased, which means there was a net gain of energy in the system.