# Find the magnitude of the acceleration acm of the center of mass of the spherical shell.

A hollow spherical shell with mass 1.80kg rolls without slipping down a slope that makes an angle of 34.0∘ wit?

A hollow spherical shell with mass 1.80kg rolls without slipping down a slope that makes an angle of 34.0∘ with the horizontal.

Part A

Find the magnitude of the acceleration acm of the center of mass of the spherical shell.

Take the free-fall acceleration to be g = 9.80m/s2 .

Part B

Find the magnitude of the frictional force acting on the spherical shell.

Take the free-fall acceleration to be g = 9.80m/s2 .

Part C

Find the minimum coefficient of friction μ needed to prevent the spherical shell from slipping as it rolls down the slope.

you were absolutely right thank you for explaining it like that too. it was easy to follow. Would you happen to know how to solve for this as well?

Two forces act along the line of motion. Gravity down the slope, m g sin(α), and friction up the slope, F. They provide the linear acceleration and F also provides the torque about the CoM. This gives the equations of motion:

m a = m g sin(α) – F

I dω/dt = F R

A spherical shell of mass m has moment of inertia I = 2/3 m R^2. Furthermore a pure rolling relates dω/dt and a through a = R dω/dt. So the two equations become

m a = m g sin(α) – F

2/3 m a = F

These we solve for a and F:

Substituting F from the second equation gives

m a = m g sin(α) – 2/3 m a

a = 3/5 g sin(α)

——————

For part A this gives

a = 3/5 * 9.80m/s^2*sin(34.0)= 3.29 m/s^2

——————-

With that we can substitute back to find F:

F = 2/3 m 3/5 g sin(α)

F = 2/5 mg sin(α)

——————–

For part B this gives

F = 2/5*1.80kg*9.80N/kg*sin(34.0) = 3.95N

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