a merry-go-round of radius 2 m and a moment of inertia 651 kgmm is stationary. a 78 kg woman runs at 1.6 m/s and jumps on tangentially to the rim of the merry-go-round. what is the final angular velocity of the merry-go-round??
0.383 rad/s? 17.672 rad/s? 0.554 rad/s? 0.259 rad/s? 0.309 rad/s?
final angular momentum about axis of rotation = initial angular momentum
initial ang mom = 78*1.6 * 2
= 250 kg m^2/s
final ang mom = 78 (2)^2 omega + 651 omega
omega ( 963) = 250
omega = .26 radians/second
To find the final angular velocity of the merry-go-round after the woman jumps on, we can use the principle of conservation of angular momentum. The initial angular momentum of the system (merry-go-round + woman) is equal to the final angular momentum.
The initial angular momentum is given by the equation:
L_initial = I_initial * ω_initial
Where:
L_initial is the initial angular momentum,
I_initial is the initial moment of inertia of the merry-go-round,
ω_initial is the initial angular velocity.
Since the merry-go-round is initially stationary, the initial angular velocity is 0.
L_initial = I_initial * 0
= 0
After the woman jumps on, the system starts to rotate with a final angular velocity (ω_final). The total angular momentum is given by:
L_final = I_final * ω_final
Where:
L_final is the final angular momentum,
I_final is the final moment of inertia of the system (merry-go-round + woman),
ω_final is the final angular velocity we want to find.
To calculate the final moment of inertia (I_final), we need to consider the moment of inertia of the merry-go-round (I_mg) and the woman (I_w) separately.
I_final = I_mg + I_w
The moment of inertia of a point mass rotating at a distance (r) from the axis of rotation is given by:
I = m * r^2
For the merry-go-round, given a radius (r_mg) of 2 m and a moment of inertia (I_mg) of 651 kgmm, we can calculate:
I_mg = m_mg * r_mg^2
Similarly, for the woman, given her mass (m_w) of 78 kg and the fact that she jumps on tangentially to the rim, we can calculate her moment of inertia (I_w):
I_w = m_w * r_mg^2
Substituting these values into the equation for I_final:
I_final = I_mg + I_w
We can now substitute the values into the equation for the final angular momentum:
L_final = I_final * ω_final
Since the initial angular momentum is 0 and the final angular momentum (L_final) is equal to the product of the final moment of inertia (I_final) and ω_final, we have:
L_final = I_final * ω_final = 0
Therefore, the final angular velocity (ω_final) of the merry-go-round is 0. The correct answer is not among the options provided.