Answer:
a) 11.7 kg. m2/s b) 0.76 Kg. m2 c) -0.33 N.m
Explanation:
a)
- Assuming no external torques act on the skater, total angular momentum must be conserved:
L1 = L2
As the angular momentum can be calculated as the
product of the moment of inertia times the angular velocity,
we can write:
I1*ω1 = I2*ω2
The initial angular momentum can be written as follows:
I1*ω1= 0.31 kg.m2 * 6.0 rev/sec
As we need to express the angular momentum in kg.m2/s, we need to convert the angular velocity units, from rev/sec to rad/sec, as follows:
ω1= 6.0 rev/sec (2π rad/ rev) = 12 π rad/sec
I1*ω1= 0.31 kg.m2 * 12 π rad/sec = 11.7 kg. m2/s
b)
- As the final angular momentum must be the same, and we know the value of the final angular velocity, we can replace by the values in L2, and solve for I2, as follows:
I2 = I1*( ω1 / ω2) = 0.31 kg. m2 . 6.0/2.45 = 0.76 kg.m2
c)
- If an external torque is present, we can write the following equation, that relates the external torque with the rotational inertia and the angular acceleration, as follows:
Τ= I *γ (1)
Where γ, is the angular acceleration.
By definition, γ is the rate of change of the angular velocity,
so if we have the values of the initial and final angular
velocities, and the time passed, we can express γ as
follows:
γ= (ω2 – ω1) / t
In order to express γ in rad/sec2, we need to convert the
angular velocities (given in rev/sec), to rad/sec, as follows:
ω1= 6.0 rev/sec (2π rad/ rev) = 12 π rad/sec
ω2= 3.0 rev/sec (2π rad/ rev) = 6 π rad/sec
Solving for γ:
γ = -6 π / 18. 0 rad/sec2 = -1.05 rad/sec2
Replacing in (1), we have:
τ= 0.31 kg. m2.*(-1.05 rad/sec2) = -0.33 N.m
