during a throw an athlete exerts a net average torque of 123Nm on a discus about its axis of spin. the mass of the discus is 2.2kg and its radius of gyration about its spin axis is 14cm. if the discus was not spinning at the beginning of the throw and the throwing action lasts for 0.6s, calculate how fast (in radians s) the discus is spinning when it is released.
To calculate the final angular velocity of the discus when it is released, we can use the principle of conservation of angular momentum.
The initial angular momentum of the discus is 0, since it is not spinning at the beginning of the throw. The final angular momentum can be calculated using the formula:
L = I*w
Where:
L = angular momentum
I = moment of inertia
w = angular velocity
The moment of inertia of the discus about its spin axis can be calculated using the formula for the moment of inertia of a disk:
I = (1/2)*m*r^2
Where:
m = mass of the discus
r = radius of gyration
Plugging in the values:
I = (1/2)*2.2kg*(0.14m)^2 = 0.02156 kg*m^2
The average torque exerted on the discus during the throw is:
τ = I*α
Where:
τ = torque
α = angular acceleration
The angular acceleration can be calculated as:
α = τ/I = 123Nm / 0.02156kg*m^2 = 5707.8 rad/s^2
Using the formula for angular velocity:
ω = ω_0 + α*t
Where:
ω = final angular velocity
ω_0 = initial angular velocity = 0
t = time of the throw = 0.6s
Therefore, the final angular velocity of the discus when it is released is:
ω = 0 + 5707.8 rad/s^2 * 0.6 s = 3424.68 rad/s
So, the discus is spinning at a speed of 3424.68 radians per second when it is released.