The average speed is 52km
<span>Assuming that the momenta of the two pieces are equal: when they have equal velocities, then
the masses of the two pieces are also equal.
Since there is no force from outside of the system, the center of mass moves on with the same velocity as before the equation. So the two pieces must fly at the side side of the mass center, i.e., they must always be at 90° to the side of the mass center. Otherwise it would not be the mass center, respectively the pieces would not have equal velocities.
This is only possible, when the angle of their velocity with the initial direction is 60°.
Because, cos (60°) = 1/2 = v/(2v).</span>
At the bottom of the rotation, the kinetic far exceeds the potential. However, at both tops, potential exceeds kinetic.
Answer:
magnitude of the frictional torque is 0.11 Nm
Explanation:
Moment of inertia I = 0.33 kg⋅m2
Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s
Final angular velocity w = 0 (since it stops)
Time t = 13 secs
Using w = w° + §t
Where § is angular acceleration
O = 4.34 + 13§
§ = -4.34/13 = -0.33 rad/s2
The negative sign implies it's a negative acceleration.
Frictional torque that brought it to rest must be equal to the original torque.
Torqu = I x §
T = 0.33 x 0.33 = 0.11 Nm
