In trigonometric functions, the period is always given by , when the function is written as . Once, we determine our  value, we are halfway to the solution!

Using conservation of energy:

Plugging in values:

Solving for 

Plugging in values:

In order to find the distance the wheel travels, we need a way to convert angular displacement to linear displacement. We know that the circumference of a circle (or a wheel, in this case) is . This means that in one rotation, the wheel would travel a distance equal to its circumference.

Distance of one rotation equals: .

Since the wheel travels  rotations, the total distance that the wheel travels will be equal to the distance traveled by one rotation multiplied by the number of rotations.

Distance of rotations equals:

The equation for centripetal force on an object moving at a constant speed in a circular path is

The weight of the person is given to be .  Assuming , we can find the mass of the person using the equation

We are given the velocity, , and the radius, .  So, we can now solve for the centripetal force:

The faster that the object is moving while being spun, the greater the force of tension in the rope will be. Therefore, if we want to find the minimum speed of the bucket at the highest point in the circle such that the rope doesn't go slack, we should assume that the tension at that point is zero.

If the tension is zero, then the only force acting on the bucket at that point would be the force of gravity pulling down on the bucket. If the rope does not go slack, then the bucket must still be moving in a circular motion, so we can use the equation for circular/centripetal motion:

Since gravity is the only force, then

We can divide by  on both sides to cancel that out, and then solve for :

The radius was given to us as 55 cm. We have to convert this to meters. Since , we should divide by 100 to get .  Now, plug that into the equation above and solve for velocity:

The formula for centripetal acceleration is:

First, we need to solve for velocity by taking the distance Bob travels and dividing by the time it takes to travel that distance. The distance he travels is the circumference of the circle, and the time is 30 seconds, because the merry-go-round completes two revolutions per minute, or one revolution every thirty seconds:

Therefore,

The ring has more of it's mass further from the center than the disk, which is more spread out. Moment of inertia increases if there is a larger "circle" that the mass has to make. Thus, if they have the same total mass, the disk will have a lower moment of inertia, thus it will reach the bottom faster. An example of this phenomenon is that of an ice skater holding in his/her arms and legs close to the center of rotation when doing jumping spins. If the ice skater were to spread his/her limbs out, their moment of inertia increases, and they would not be able to spin as fast.

For this question, we have to consider the case of the earth orbiting the sun. Just like any scenario in which an object is moving in a circular motion, we know that there is a centripetal force that is constantly pulling the object inward. Thus, even though the object's speed may be constant, its velocity is constantly changing. This is because speed is a scalar that is not affected by direction. Velocity, on the other hand, is a vector which depends on direction.

To answer this question, we'll need to first identify all the forces acting on the hanging mass. To do this, we have to consider the forces acting along the y-axis and x-axis.

First, let's consider the forces in the y-direction. Since the mass is not moving in the y-direction, we know that the net force acting on the mass in the vertical direction is zero. We know that the weight of the mass acts downward, and the vertical component of the string's tension acts upward.

Next, let's look at the forces acting on the mass in the horizontal direction. Since the mass is moving in a uniform circular motion, we know that there is a centripetal force associated with this. Moreover, we know that the x-direction of the tension in the string is acting in the x-direction. Thus, this component of the string's tension must be providing the centripetal force.

Now, if we divide these two equations, we can cancel out the terms for tension and mass.

Further rearrangement of the above expression allows us to solve for the angle.

For this question, we're asked to evaluate a true statement regarding circular motion. Hence, we'll need to look at each answer choice individually and consider whether it is true.

• The centripetal acceleration provided by the Earth's gravitational field is constant

This is not a true statement. Remember that acceleration is a vector. Even though the astronaut's centripetal acceleration is constant in magnitude, it is constantly changing direction as the force will always point towards the center of the Earth.

• If the astronaut's speed is doubled, his centripetal acceleration will also double

This statement is false. Remember the expression for centripetal force.

Hence, if the astronaut's speed becomes twice as great, then his acceleration will become four times as great.

• If the distance between the astronaut and the Earth were to double, the centripetal acceleration of the astronaut would also double

This statement is not true. Again, we need only look to the expression shown above for centripetal force to see why. Because the centripetal acceleration and distance are both inversely proportional to one another, doubling the distance of the astronaut from the Earth would cause the acceleration to become half as great. Thus, it would decrease, not increase.

• The astronaut is constantly in a state of free-fall

This is a true statement, and thus it is the correct answer. Objects that are orbiting the Earth experience a centripetal force due to gravity. If we picture a snapshot of the movement of the astronaut, he will be moving tangentially to his circular displacement. As the astronaut moves tangentially, the force of gravity serves as the centripetal force that "pulls" him back, such that the astronaut does not move down the tangential path. This continuous "pull" away from the tangential path at each instant essentially means that our astronaut will constantly fall towards the earth while maintaining his circular path, and is thus constantly in a state of free-fall, where he will be weightless.