We can solve this problem by looking at all of the forces acting on the car at the top of the circular path.

The weight is positive, because it points towards the center of the circle, while the normal force  is negative because it points away from the circle.

Since we are asked to look at the special case where the occupants of the car feel weightless, this tells us that the normal force experienced by the occupants must be zero.

For this problem, we recognize that the force keeping the mass m suspended is the tension in the string. This tension is balanced by the gravitational force on m such that m does not accelerate. So how do we relate the speed of M to the tension? We must ask ourselves why is there tension in the string at all? Any object that travels in a circle has some centripital acceleration acting on it that point towards the center of the circle (for constant speed). This force is equal to the mass time the centripial acceleration. 

I have stated previously that this force must be equal to the gravititational force acting on mass m. So,

Solving for v,

Plugging in values yields a speed of 1.6 m/s

Let's start with what we know, and what we want to know. We are given the mass of an object, and we're told that it's traveling in a circular orbit. Also, we have been given the diameter of the orbit. What we are trying to find is the velocity of the object.

In this problem, we'll need to consider the centripetal force of the object as it travels in a circular orbit. The force that is contributing to the centripetal force is the weight of the object. Therefore, we can write the following expression:

The above expression tells us that the object's velocity is dependent on the acceleration due to gravity, as well as the radius of the orbit. Since we're told that the diameter of the orbit is , we can find the radius by taking half of this value, which is . Moreover, the plus-or-minus sign indicates that the object can either be traveling in a clockwise or counter-clockwise orbit. In either case, the object will have the same magnitude of velocity.

Solving for velocity, we can plug in values to obtain:

Because the hammer is moving at a constant angular velocity, the centripetal force along the hammer rope can be expressed as . By plugging in the numbers given by the problem we get  which, when simplified gives us approximately  or 

In order to find the radius of the circle that corresponds to the maximum tension the string can experience without breaking, we must rearrange the centripetal force equation for objects moving in a circle:

To solve this problem, it is important to remember the signs of each force acting on the coaster.  Since we are inside the loop at the top, both the weight and the normal force point toward to center of the circular path, and are thus positive.

Next, we note that the normal force is what a rider "feels" during their ride on the rollercoaster.  We want the speed at which a rider would feel three times as heavy, or three times their weight.

Centripetal acceleration  can be given as:

, where  is the linear velocity,  is the radius of the circle. 

To determine linear velocity, we need to determine how quickly the wheel spins.

Firstly, let's convert from rpm to Hertz. 

The wheel makes 50 full revolutions in one second. To determine linear velocity,

, where  is frequency, which we just determined and  is circumference of the tire. 

By the definition of the circumference of a circle,

Therefore, 

The formula for centripetal acceleration is given by:

 is the centripetal acceleration,  is the velocity and  is the radius of the circular path. We substitute in our known values and then solve for alpha which will give us the centripetal acceleration of the ferris wheel.

Be aware that the radius is half of the diameter, so  will be 5 meters.

This will give us a centripetal acceleration of 

The tension and gravity both point down toward the center of the circle. The net force on the ball is the vector sum of these two: 

For an object in circular motion, the net force must be equal to:

In this case, the force of gravity is helping the tension provide this force, so the tension is not as great as it would be if it were supplying the entire centripetal force.

Substitute and solve.

For the car going around a flat turn, the force of friction is the only force keeping the car in a circular path. This lets us equate the centripetal force to the force of friction.

Substitute and solve for the coefficient of friction.

Since the tires are rotating but not slipping on the surface, this is the coefficient of static friction, .