A merry-go-round with a radius of 6.0 m isset rotating from rest in a clockwisedirection and attains an angular speed of0.50 −1rad s in 100 s. What is themagnitude and direction of the torqueexperienced by a child of mass 30 kg aboutthe centre of the merry-go-round if she issitting at a distance of 3.0 m from thecentre ?What is her rotational kineticenergy ?
Question
A merry-go-round with a radius of 6.0 m isset rotating from rest in a clockwisedirection and attains an angular speed of0.50 −1rad s in 100 s. What is themagnitude and direction of the torqueexperienced by a child of mass 30 kg aboutthe centre of the merry-go-round if she issitting at a distance of 3.0 m from thecentre ?What is her rotational kineticenergy ?
Solution
To find the magnitude and direction of the torque experienced by the child, we can use the formula:
Torque = Moment of Inertia * Angular Acceleration
First, let's find the moment of inertia of the merry-go-round. The moment of inertia for a solid disk rotating about its center is given by the formula:
Moment of Inertia = (1/2) * Mass * Radius^2
Plugging in the values, we get:
Moment of Inertia = (1/2) * (30 kg) * (6.0 m)^2 = 540 kg·m^2
Next, we need to find the angular acceleration. The angular acceleration is the change in angular speed divided by the time taken. In this case, the change in angular speed is 0.50 rad/s and the time taken is 100 s. So,
Angular Acceleration = (0.50 rad/s) / (100 s) = 0.005 rad/s^2
Now we can calculate the torque:
Torque = (540 kg·m^2) * (0.005 rad/s^2) = 2.7 N·m
The magnitude of the torque experienced by the child is 2.7 N·m.
To determine the direction of the torque, we can use the right-hand rule. If the child is sitting at a distance of 3.0 m from the center and the merry-go-round is rotating clockwise, then the torque will be directed outwards from the center of the merry-go-round.
Next, let's calculate the rotational kinetic energy of the child. The rotational kinetic energy is given by the formula:
Rotational Kinetic Energy = (1/2) * Moment of Inertia * Angular Speed^2
Plugging in the values, we get:
Rotational Kinetic Energy = (1/2) * (540 kg·m^2) * (0.50 rad/s)^2 = 67.5 J
The rotational kinetic energy of the child is 67.5 Joules.
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