# Merry go round torque

There is no friction on the axle of the merry-go- round, so, because it equals zero, friction on the axle does not cause an external torque on the merry-go-.For example, if you have a merry-go-round, you can apply a force to the rim of it. If the force is towards the center, it won’t move at all. figure57.Expert Answers: It is ok, the net torque on the merry go round is zero (which really should be a vector). This means that the angular. If the system is the bullet and the merry go round, the net torque is zero. This means that the change in angular momentum is zero or that the. The torque in a case will be zero if the angular momentum is constant. There is no torque since you just jumped off the merry-go-round. On the off chance that.

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## merry-go-round physics problem

Why is there no torque on the merry go round? There is no torque because you just stepped off. If you had jumped off, it could make a difference. The problem is ignoring the vertical component because the platform is constrained to rotate in the horizontal plane. Any (vertical) torque from. The merry go round can turn, but its center of mass can’t move. When the bullet hits, the axle exerts a force to prevent the merry go round from. Sarah and the merry-go-round (mass M, radius R, and I = cMR2) then spin together with angular velocity ?f. If Sarah’s initial velocity is tangent to the. The child can be considered as a point mass. a) What is the new angular velocity of the merry-go-round? b) Does the total kinetic energy of the system.

## angular speed merry-go-round

The same as before. So, although the person is moving in a straight line the angular momentum (about that rotation point) is constant. The total. No, you are going faster. The merry-go-round is rotating with uniform circular motion – that means that it rotates at a constant angular speed.Do you remember playing on a merry-go-round when you were younger? [Figure 1]. If two people are riding on the outer edge, their velocities. For the angular momentum principle, the change in angular momentum is equal to the net torque. If the system is the bullet and the merry go. Solution. (a) Find the angular speed when the student reaches a point 0.500 m from the center. Calculate the moment of inertia of the disk, Iñi. ID=¹/MR² = ¹(.

## jumping off a merry-go-round physics

Physics: Suppose a man is standing on a merry-go-round which is rotating with some speed, if he jumps out in the radial direction, then the rotation speed. Download Citation | Jumping Off a Merry-Go-Round | How many vertical jumps does it take to jump off of a rotating merry-go-round (MGR)?T Toepker · 2022  How many vertical jumps does it take to jump off of a rotating merry-go-round (MGR)? The answer is hidden in the expression: rn = r0 n/2.Treat the merry go round as a solid disc, and Fred as a point object. As it turns, he walks to the edge of the merry-go-round, then jumps o?. It’s worth noting that the “jump force” ?J will in fact exert a torque on the merry-go-round, due to its upward component. But this torque.

## merry-go-round physics problem angular momentum

angular momentum when the student has reached r = 0.500 m. The key is to find the different moments of inertia. M. Ad m. Page 2. 8.7 Angular Momentum.The same as before. So, although the person is moving in a straight line the angular momentum (about that rotation point) is constant. The total. The problem is ignoring the vertical component because the platform is constrained to rotate in the horizontal plane. Any (vertical) torque from. For the angular momentum principle, the change in angular momentum is equal to the net torque. If the system is the bullet and the merry go. Apply conservation of angular momentum (there are no net external torques on the system of merry-go-round and child). Thus we have: L = constant = Ii i = If f.