You wish to accelerate a small merry-go-round from rest to a rotational speed of one-fourth of a revolution per second by pushing tangentially on it. Assume the merry-go-round is a disk with a mass of 300 kg and a radius of 2.00 m .
Ignoring friction, how hard do you have to push tangentially to accomplish this in 6.00 s ? (Use energy methods and assume a constant push on your part.)
A person opens a door by applying a 13-N force perpendicular to it at a distance 0.70 m from the hinges. The door is pushed wide open (to 120 ? ) in 2.3 s .
How much work was done?
What was the average power delivered?
First Question -
(1) First convert the units of required rotational speed into radian/sec –
w = 0.25 x 2 x pi = 1.5708 rad/s
given that, requisite time to accomplish the speed, t = 6
So, the required rotational acceleration in rad/s^2:
a = w / t = 1.5708 / 6 = 0.262 rad/s^2
Moment of inertia for the disc, I = 0.5 x m x r^2 = 0.5 x 300 x 2^2 = 600 kgm^2
So, the torque required to accelerate disc -
T = I x a = 600 x 0.262 = 157.2 Nm
Therefore, the requisite force -
F = T / r = 157.2 / 2 = 78.6 N
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