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# Acceleration, Friction Force, Coefficient of Friction

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A hollow, spherical shell with mass 2.00 kg rolls without slipping down a slope angled at 34.0 degrees

A) Find the acceleration (take free fall to be 9.8 m/s^2)

B) Find the friction force

C) Find the minimum coefficient of friction needed to prevent slipping

https://brainmass.com/physics/torques/acceleration-friction-force-coefficient-friction-352719

## SOLUTION This solution is FREE courtesy of BrainMass!

Since the sphere doesn't slip;

torque = (moment of inertia of sphere)(angular acceleration) = (force due to friction)(radius of sphere)

I&#945; = (Ff)(r)

where;

I=2mr^2/3
&#945;=a/r

if we substitute and solve for Ff;
(2mr^2/3)(a/r) = (Ff)(r)
Ff = 2ma/3

Net force acting on the system is;

Fnet=(force due to gravity along slope) - (force due to friction)

Fnet = (force due to gravity along slope) - (force due to friction)

Fnet = mgsin&#952; - Ff

Fnet = mgsin&#952; - 2ma/3 = ma

a=gsin&#952; - 2a/3

a=3gsin&#952;/5

sin 34=0.56

a=3x9.81x0.56/5

a=3.29m/s2

Now we can calculate Ff;

Ff=2ma/3=2x2x3.29/3

Ff=4.38N

Minimum friction coefficient for the non-slip condition can be calculated from;

Ff = &#956;mgcos&#952;

cos 34=0.83

4.38=&#956;x2x9.81x0.83

&#956;=4.38/16.28

&#956;=0.27

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!