A physics major is working to pay his college tuition by performing in a traveling carnival. He rides a motorcycle inside a hollow transparent plastic sphere. After gaining sufficient speed, he travels in a vertical circle with a radius of 14.6 m . The physics major has a mass of 72.0 kg , and his motorcycle has a mass of 40.0 kg.(a) What minimum speed must she have at the top of the circle for the motorcycle tires to remain in contact with the sphere? (b) At the bottom of the circle, her speed is twice the value calculated in part (a). What is the magnitude of the normal force exerted on the motorcycle by the sphere at this point?

A) 11.968 m/s

B) 2668.4N

Explanation:

Mass of motorcycle (Mm) = 40kg

Mass of Physics major (Mp) = 72kg

Radius (r) = 14.6m

A) Using Newton's second law of motion, the net force at the top of circle is given by;

ΣF = - Ma(c) = -Mv²/r

And so; Resultant(R) = Mg - Ma(c)

Now, R= 0 since it loses contact.

Thus,

Mg - Ma(c) = 0

From above, -Ma(c) = -mv²/r

So we'll get; Mg - M(vmin)²/r = 0

Thus,Mg = M(vmin)²/r

So, g = (vmin)²/r

(vmin)² = rg

Vmin = √rg

Vmin = √(14.6 x 9.81)

Vmin = √143.226 = 11.968 m/s

B) From the earlier equation, the net force acting on the motorcycle is;

R(m) - M(m)g - M(p)g = M(m)a(c) = (M(m)v²)/r

Thus; let's use;

R(m) - M(m)g - M(p)g = (M(m)v²)/r

So;

R(m) = M(m)g + M(p)g + (M(m)v²)/r

R(m) = g[M(m) + M(p)] + (M(m)v²)/r

From the question, v is now double.

So, v = 2 x 11.968 = 23.936

R(m) = 9.81[40 + 72] + [(40 x 23

936²)/14.6]

R(m) = 1098.72 + 1569.68 = 2668.4N

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