Question

An object of mass m = 3.65kg rolls off of an incline plane from a height of 8.75m, where the angle of the incline measured from the horizontal is 31.3o. If the object is a solid uniform cylinder of radius r = 1.55m, then at the bottom of the ramp, what is the (a) rotational kinetic energy and (b) the translational kinetic energy. If the object is a hoop of radius, r = 1.15m, then at the bottom of the ramp, what is the (c) rotational kinetic energy and (d) the translational kinetic energy?

Answer 1

a)

h = height dropped = 8.75 m

m = mass of the object = 3.65 kg

r = radius = 1.55 m

Using conservation of energy

Potential energy at the top = rotational kinetic energy + translational kinetic energy at bottom

mgh = (0.5) I w² + (0.5) m v²

for solid cylinder : I = (0.5) m r² and w = v/r

hence

mgh = (0.5) (0.5) (m r²) (v/r)² + (0.5) m v²

mgh = (0.25) m v² + (0.5) m v²

mgh = 0.75 m v²

(9.8) (8.75) = (0.75) v²

v = 10.7 m/s

w = angular speed = v/r = 10.7 /1.55 = 6.9 rad/s

I = moment of inertia = (0.5) m r² = (0.5) (3.65) (1.55)² = 4.4 kgm²

rotational kinetic energy is given as

RKE = (0.5) I w²

RKE = (0.5) (4.4) (6.9)²

RKE = 104.74 J

b)

translational Kinetic energy is given as

KE = (0.5) m v² = (0.5) (3.65) (10.7)² = 209 J

c)

h = height dropped = 8.75 m

m = mass of the object = 3.65 kg

r = radius = 1.15 m

Using conservation of energy

Potential energy at the top = rotational kinetic energy + translational kinetic energy at bottom

mgh = (0.5) I w² + (0.5) m v²

for a hoop : I = m r² and w = v/r

hence

mgh = (0.5) (m r²) (v/r)² + (0.5) m v²

mgh = (0.5) m v² + (0.5) m v²

mgh = m v²

(9.8) (8.75) = v²

v = 9.3 m/s

w = angular speed = v/r = 9.3 /1.15 = 8.1 rad/s

I = moment of inertia = (0.5) m r² = (0.5) (3.65) (1.15)² = 2.41 kgm²

rotational kinetic energy is given as

RKE = (0.5) I w²

RKE = (0.5) (2.41) (8.1)²

RKE = 79.1 J

d)

translational Kinetic energy is given as

KE = (0.5) m v² = (0.5) (3.65) (9.3)² = 157.8 J