During a throw an athlete exerts a net average torque of 100 Nm on a discus
about its axis of spin. The mass of the discus is 2 kg and its radius of gyration
about its spin axis is 12 cm. If the discus was not spinning at the beginning of
the throw and the throwing action lasts for 0.2 s, calculate how fast (in
radians s-1) the discus is spinning when it is released. [8 marks]
694.44rad/s
To find the angular velocity of the discus when it is released, we can use the principle of conservation of angular momentum.
The moment of inertia of the discus is given by the formula:
I = m * r^2
Where:
I = moment of inertia
m = mass of the discus
r = radius of gyration
Substituting the given values, we have:
I = 2 kg * (0.12 m)^2
I = 0.288 kgm^2
The initial angular momentum is zero since the discus is not spinning at the beginning. The final angular momentum is given by the formula:
L = I * ω
Where:
L = angular momentum
I = moment of inertia
ω = angular velocity
The net average torque can be calculated using the formula:
τ = I * α
Where:
τ = torque
I = moment of inertia
α = angular acceleration
Rearranging the formula, we can solve for α:
α = τ / I
Substituting the given values:
α = 100 Nm / 0.288 kgm^2
α ≈ 347.22 rad/s^2
Since the throwing action lasts for 0.2 seconds, we can find the final angular velocity using the formula:
ω = α * t
Substituting the values:
ω = 347.22 rad/s^2 * 0.2 s
ω ≈ 69.44 rad/s
Therefore, the discus is spinning at approximately 69.44 radians per second when it is released.