Question

A hollow spherical shell with mass 2.50 kg rolls without slipping down a slope that makes an angle of 33.0 degrees with the horizontal.

If g=-9.8m/s^2, find the magnitude of the acceleration of the center of mass of the spherical shell and the magnitude of the frictional force acting on the spherical shell.

Answer 1

NOTE - Is it a negative sign in front of g value ??? or is it by mistake ??? please verify it..

NOTE - Is it a negative sign in front of g value ??? or is it by mistake ??? please verify it..

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The component of weight parallel to the incline is given as

mgsin

Also,

the torque due to friction force, f is

f * R = I

where R is radius of sphere, is angular acceleration and I is moment of inertia which is (2/3)mR² for hollow sphere.

we also know that

= a / R

where a is linear acceleration.

Therefore, we have

f * R = Ia / R

f = Ia / R²

Now, use Newton's second law

mgsin - f = ma

mgsin - Ia / R² = ma

mgsin = ma + Ia / R²

mgsin = a ( m + I / R²)

a = mr²gsin / mR² + I

Now put I = (2/3)mR² , we get

a = 3/5 * g * sin

a = 3/5 * 9.8 * sin 33

a = 3.2025 m/s²

( If you want to use g = - 9.8, then a = - 3.2025 m/s²)

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Now, to calculate friction force, follow the same procedure,

f = 2/5 * m * g * sin

f = 2/5 * 2.50 * 9.8 * sin 33

f = 5.337 N