You stand on a frictionless platform that is rotating with an angular speed of 3.9 rev/s.Your arms are outstretched, and you hold a heavy weight in each hand. The moment of
inertia of you, the extended weights, and the platform is 7 kg · m2. When you pull the weights in toward your body, the moment of
inertia decreases to 5.2 kg · m2.
a)What is the resulting angular speed of the platform?
Answer in units of rev/s.
b)What is the change in kinetic energy of the system?
Answer in units of J.
a) I omegai = I omegaf
b) Do 1/2 I omegai^2 and 1/2 I omegaf^2 and compare
To find the resulting angular speed of the platform, we can use the principle of conservation of angular momentum. The angular momentum of the system remains constant when there are no external torques acting on it.
Angular momentum (L) = moment of inertia (I) × angular velocity (ω)
Initially, when the moment of inertia is 7 kg · m^2 and the angular speed is 3.9 rev/s, the initial angular momentum is:
L1 = I1 × ω1 = 7 kg · m^2 × 3.9 rev/s
When the moment of inertia decreases to 5.2 kg · m^2 (due to the weights being pulled in), the angular momentum remains constant and is equal to the initial angular momentum:
L1 = L2
I1 × ω1 = I2 × ω2
Rearranging for the resulting angular speed (ω2):
ω2 = (I1 × ω1) / I2
Substituting the given values:
ω2 = (7 kg · m^2 × 3.9 rev/s) / 5.2 kg · m^2
Simplifying:
ω2 = 5.25 rev/s
So, the resulting angular speed of the platform is 5.25 rev/s.
To find the change in kinetic energy of the system, we can use the equation:
ΔKE = KE2 - KE1
Since the system is rotating, the kinetic energy is given by:
KE = (1/2) × I × ω^2
Therefore, the change in kinetic energy is:
ΔKE = (1/2) × (I2 × ω2^2) - (1/2) × (I1 × ω1^2)
Substituting the given values:
ΔKE = (1/2) × (5.2 kg · m^2 × (5.25 rev/s)^2) - (1/2) × (7 kg · m^2 × (3.9 rev/s)^2)
Simplifying:
ΔKE = 44.59 J
So, the change in kinetic energy of the system is 44.59 J.
To solve this problem, we need to apply the law of conservation of angular momentum and kinetic energy.
a) The law of conservation of angular momentum states that the initial angular momentum of a system remains constant unless acted upon by an external torque. In this case, the platform and weights are initially rotating with an angular speed of 3.9 rev/s.
Angular momentum is given by the product of moment of inertia (I) and angular speed (ω). So, the initial angular momentum (L1) is:
L1 = I1 * ω1
When the weights are pulled in towards the body, the moment of inertia decreases to 5.2 kg · m^2 (I2). However, the angular momentum (L2) still remains constant:
L2 = I2 * ω2
Since angular momentum is conserved, we can equate the initial and final angular momentum:
L1 = L2
I1 * ω1 = I2 * ω2
Rearranging the equation, we can solve for the final angular speed (ω2):
ω2 = (I1 * ω1) / I2
Substituting the given values:
ω2 = (7 * 3.9) / 5.2
ω2 = 5.25 rev/s
Therefore, the resulting angular speed of the platform is 5.25 rev/s.
b) Kinetic energy is given by the equation:
KE = (1/2) * I * ω^2
The change in kinetic energy (ΔKE) is the difference between the initial and final kinetic energy:
ΔKE = KE2 - KE1
Substituting the values, we get:
ΔKE = (1/2) * I2 * ω2^2 - (1/2) * I1 * ω1^2
Using the calculated values from part a, we can calculate the change in kinetic energy:
ΔKE = (1/2) * 5.2 * (5.25^2) - (1/2) * 7 * (3.9^2)
Simplifying the equation:
ΔKE = 34.175 J
Therefore, the change in kinetic energy of the system is 34.175 J.