Similar to linear acceleration, angular acceleration is related to the angular velocity by the following

from here we simply plug in our values of  and  to get . Next to convert to radians we take the relations that pi is equal to  rotation and multiply our  by  to arrive at our final answer of 

Most of the success for this ride experience can be credited to Newton's Third Law. It states that if object 1 exerts  onto object 2, then object 2 must be exerting a force . They are equal and opposite forces, called an action reaction pair. This concept applies when you're typing on your cell phone. when you touch down on your screen or button, your finger doesn't go through the phone, right? The phone is sending back an equal force to you which prevents your finger from going through the screen.

The same thing applies on the attraction. Your body is pushing outward on the walls of the ride. If no centripetal force was present, your body would project on a tangent as depicted in the diagram below. However, centripetal force is present. This is the force that is countering your body; it is pushing you towards the center. Now as we know, centripetal force isn't ever drawn on a diagram because it is a "net" force, or a collection of forces. But what we do know is that during circular motion, centripetal force (and acceleration) point towards the center.

During the ride, your body is trying to follow the tangental paths shown, but the reciprocal force, , is keeping your body stationary on the wall. The wall is following a circular path, which makes you follow the circular path. This action reaction pair is what allows the attraction to work. Centripetal force points inwards.

The minimum length of the wrench will assume that the maximum force is applied at an angle of . Therefore, we can use the simplified expression for torque:

Here,  is the length of the wrench.

Rearranging for length and plugging in our values, we get:

The counterclockwise and clockwise torques about the pivot point must be equal for the rod to balance. Taking the fulcrum as the pivot point, the counterclockwise torque is due to the rod’s weight, gravitational force acting downwards at the center of the rod. If we use the pivot as our reference, then the center of the rod is 15cm from the reference.

Set this equal to the clockwise torque due to the additional mass, a distance r to the right of the pivot.

.

Solving for r gives r = 0.05m to the right of the pivot, so 40 + 5 cm from the left end of the rod.

The net torque on the pulley is zero. Remember that , assuming the force acts perpendicular to the radius. Because the pulley is symmetrical in this problem (meaning the r is the same) and the tension throughout the entire rope is the same (meaning F is the same), we know that the counterclockwise torque cancels out the clockwise torque, thus, the net torque is zero.

In the image below, T₁ (due to the platform with the 4 0.5kg weights) = T₂ (the 2kg mass).

Torque is defined by the equation . Since both students will exert a downward force perpendicular to the length of the seesaw, . In our case, force is the force of gravity, given below, and is the distance from the center of the seesaw.

Since the torque must be zero in order for the seesaw to stay parallel (not move), the lighter student on the right must make his torque on the right equal to the torque of the student on the left. We can determine the required distance by setting their torques equal to each other.

Torque, in this case, is dependent on both the force exerted by the students as well as their distances from the point of rotation. As a result, both students moving forward by one meter will cause a nonzero torque on the seesaw. This is because the heavier student's ratio of force and distance will result in less torque on his side than the lighter student.

This a an example of rotational equilibrium involving torque. The formula for torque is , where  is the angle that the force vector makes with the object in equilibrium and  is the distance from the fulcrum to the point of the force vector. To achieve equilibrium, our torques must be equal.

Since the forces are applied perpendicular to the beam, becomes 1. The distance of the fulcrum from the left end is 1m and its distance from the right end is 2m.

Since the 50N force is twice as far from the fulcrum as the force that must be applied on the left side, it must be half as strong as the force on the left. The force on the left can be found to be 100N.

For the seesaw to be balanced, the system must be in rotational equilibrium. For this to occur, the torque the same on both sides.

The total torque must be equal on both sides in order for the net torque to be zero.

Substitute the formula for torque into this equation.

Now we can use the given values to solve for the missing mass.

The acceleration form gravity cancels from each term.