A hollow spherical shell with mass 1.65 kg rolls without slipping down a slope that makes an angle of 38.0 ∘ 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 = 9.80 m/s2.

The acceleration is 3.62 m/s²

Explanation:

Step 1: Data given

mass of the shell = 1.65 kg

angle = 38.0 °

Step 2: Calculate the acceleration

We have 2 forces working on the line of motion:

⇒ gravity down the slope = m*g*sinα

⇒ provides the linear acceleration

⇒ friction up the slope = F

⇒ provides the linear acceleration and also the torque about the CoM.

∑F = m*a = m*g*sin (α) - F

I*dω/dt = F*R

The spherical shell with mass m has moment of inertia I=2/3*m*R² 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

IF we combine both:

m*a = m*g*sin (α) - 2/3*m*a

1.65a = 1.65*9.81 * sin (38.0) - 2/3 * 1.65a

1.65a + 1.1a = 9.9654

2.75a = 9.9654

a = 3.62 m/s²

The acceleration is 3.62 m/s²