Question

13. Points A wheel is made up of a uniform thin rim (hollow cylinder) of mass 2m kg and 6 thin uniform spokes each of mass m kg and length L = 0.5 meters. The wheel is given an initial translational speed Vo = 10.0 m/s and launches vertically from the top of a quarterpipe of height h -5.00 meters. The wheel rolls without slipping along the ramp, and air resistance is negligible. a) Find the moment of intertia about an axis that is both through the center and perpendicular to the wheel in terms of m and L. b) Find the angular velocity of the wheel after it has launched into the air (just as it reaches the top of the ramp). If you were unable to complete Part A, you can use / as the moment of inertia of the wheel and try to solve the problem algebraically. spoke

Answer 1

given :

vo = initial speed = 10 m/s

M = mass of ring = 2m kg

M ' = mass of each spoke = m kg

R = radius of ring = L = length of each spoke = 0.5 m

h = 5 m

a) Moment of inertia of the wheel about the said axis = $MR^{2} + \frac{6M'L^{2}}{3} = 2m \times L^{2}+2m\times L^{2} = 4mL^{2}kgm^{2}$ [answer]

b) Total kinetic energy of a purely rolling body on a fixed surface = $\frac{1}{2}mv^{2}(1+\frac{k^{2}}{R^{2}}) = \frac{1}{2}m(\omega R)^{2}(1+\frac{k^{2}}{R^{2}})$ [where $\omega$ = angular velocity, k = radius of gyration, here k = $L/\sqrt{2} = R/\sqrt{2}$ ]

applying conservation of mechanical energy :

$\frac{1}{2}8mv_{o}^{2}(1+\frac{k^{2}}{R^{2}}) = 8m gh + \frac{1}{2}8m(\omega R)^{2}(1+\frac{k^{2}}{R^{2}})\\ \Rightarrow v_{o}^{2}(1+\frac{1}{2}) = 2gh + (\omega R)^{2}(1+\frac{1}{2})$

=> 100(3/2) = (2x9.8x5) + $\omega^{2}$ x 0.05 x 3/2

=> $\omega^{2}$ = 693.33

=> $\omega$ = 26.33 rad/s [answer]