Question

A 42cm diameter wheel consists of a rim made from a rigid plastic, with linear density of 25.0g/cm and 6 spokes (hint: 6 slender rods of linear density), released at the top of a 58.0m. high hill.

How fast is the wheel rolling (i.e. angular velocity) at the bottom of the hill?

Please make it understandable

1) 248 rad/s

2) 124 rad/s

3) None of the above

Answer 1

Given that,

diameter = 42 cm

so, radius R = d / 2 = 0.21 m

linear density, = 25 g/cm

h = 58 m

Mass of rim, m_r = 2R

mass of spokes, m_s = 6*(R)

Total mass m =  2R + 6R = 2R ( + 3)

Total moment of inertia,l = l_r + l_s

l = m_r R^2 + 6*(1/3)m_s R^2

velocity v = w*R

From conservation of energy,

PE = KEt + KEr

mgh = (1/2)mv^2 + (1/2)lw^2

2R ( + 3) * gh = [0.5 * 2R ( + 3) * (wR)^2] + [(0.5*2R*R^2 + 6*(1/3)*6R*R^2) * w^2]

By solving this,

angular velocity w = sqrt [( + 3)*gh / R^2*( + 2)]

w = sqrt [( + 3)*9.8*58 / 0.21² * ( + 2)]

w = 124 rad/s