Question

In the Thunder Sphere, a motorcycle moves on the inside of a sphere, traveling in a horizontal circle along the equator of the sphere. The inner radius of the sphere is \(13.75 \mathrm{~m}\), and the motorcycle maintains a speed of \(17.01 \mathrm{~m} / \mathrm{s}\). What is the minimum value for the coefficient of static friction between the tires of the motorcycle and the inner surface of the sphere to ensure that the motorcycle does not fall?

Short answer

Answer: The minimum coefficient of static friction needed to keep the motorcycle from falling is approximately 0.48.

Step-by-step solution

Identify the forces acting on the motorcycle

There are two main forces acting on the motorcycle: (1) the gravitational force, which pulls the motorcycle downward, and (2) the normal force exerted by the sphere's surface on the motorcycle. The normal force is always perpendicular to the surface, so in this case, it acts along the radius of the sphere.

Calculate the gravitational force

The gravitational force acting on the motorcycle is given by the product of its mass (m) and the gravitational acceleration (g): \(F_g = m \times g\) Note that we don't know the actual mass of the motorcycle, but since it will cancel out when we divide forces later, we can leave it as a variable for now.

Calculate the centripetal force

The centripetal force acting on the motorcycle is given by the following formula, where v is the speed of the motorcycle and r is the radius of the sphere: \(F_c = \frac{m \times v^2}{r}\) Here, \(v = 17.01\mathrm{~m/s}\) and \(r = 13.75\mathrm{~m}\). Substituting these values, the centripetal force acting on the motorcycle is: \(F_c = \frac{m \times (17.01\mathrm{~m/s})^2}{13.75\mathrm{~m}}\)

Calculate the required frictional force

The frictional force (\(F_f\)) acting on the motorcycle is what prevents it from falling, and it must counterbalance the gravitational force (\(F_g\)). The frictional force is given by the product of the normal force (\(F_n\)) and the coefficient of static friction (\(\mu\)): \(F_f = \mu \times F_n\) In this case, the normal force is equal to the centripetal force acting on the motorcycle: \(F_n = F_c = \frac{m \times (17.01\mathrm{~m/s})^2}{13.75\mathrm{~m}}\)

Find the minimum value for the coefficient of static friction

As we stated earlier, the frictional force must counterbalance the gravitational force to prevent the motorcycle from falling: \(F_f = F_g\) \(\mu \times F_n = m \times g\) By substituting \(F_n = F_c\), we get: \(\mu \times \frac{m \times (17.01\mathrm{~m/s})^2}{13.75\mathrm{~m}} = m \times g\) Notice that the mass (m) is on both sides of the equation, so we can cancel it out: \(\mu \times \frac{(17.01\mathrm{~m/s})^2}{13.75\mathrm{~m}} = g\) Now, we can solve for the coefficient of static friction, \(\mu\), using the given values for the gravitational acceleration (\(g = 9.81 \mathrm{m/s^2}\)): \(\mu = \frac{9.81 \mathrm{m/s^2} \times 13.75\mathrm{~m}}{(17.01\mathrm{~m/s})^2}\) Finally, calculate the minimum value for the coefficient of static friction, \(\mu\): \(\mu \approx 0.48\) So, the minimum value for the coefficient of static friction needed to keep the motorcycle from falling is approximately 0.48.