A hollow sphere with a mass of M = 4 kg and radius of 30 cm is rolling down an i

Question

A hollow sphere with a mass of M = 4 kg and radius of 30 cm is rolling down an inclined surface (hill) with inclined angle of 37 degrees, from a height of 2 meters.

1. Find its rotational inertia.

2. Find its linear AND angular acceleration down the hill.

3. Find its linear AND angular velocity at the bottom of the hill.

4. Find the linear AND rotational kinetic energy at the bottom of the hill.

5. Find the frictional coefficient of the inclined surface, if the sphere is making pure rolling motion.

6. Find the number of revolutions the sphere rolls over along the inclined. (Convert rad to rev)

Explanation / Answer

1. Moment of inertia of hollow sphere is I = 2/3 mr^2

I = 0.5* 4* 0.3^2

I = 0.24 kgm^2

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3.Total PE = KEtrans + KE rot

mgh= 0.5mv^2 + 0.5I^2

using v =r

solving for v , gives v about COMi s = r/2

=2v/r

mgh= 0.5 I *4 v^2/r^2+ 0.5 mv^2

4*9.81*2 = (0.5* 0.24 * 4* v^2/0.3^2)+(0.5 * 4* v^2)

Linear velocity v = 3.27 m/s

As V = r

Angular velocity = v/r= 3.27/0.3 = 10.9 rad/s

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2. Use V^2 = 2 a d

d = H/sin 37 = 2/sin 37

d=3.32 m

so a = linear acceleration = v^2/2d

a =3.27^2/(2* 3.32)

a= linear acceleration =1.61 m/s^2

angular acceleration = a/r

= 1.61/0.3 = 5.36 rad/s^2

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4. Linear KE = 0.5 mv^2

Linear KE = 0.5 * 4* 3.27^2

Linear KE=42.77 J

Rotational KE = 0.5I^2

Rot KE = 0.5* 0.24 * 10.9^2

Rot KE=14.25 J

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