Question

Sophia is playing with a set of wooden toys, rolling them off the table and onto the floor. One of the toys is a small sphere with a mass of 0.025 kg and a radius of 0.022 m, and another is a small cylinder that also has a mass of 0.025 kg but a radius of 0.013 m. She rolls each toy so that it has the same translational speed of 0.42 m/s. How much greater is the kinetic energy of the cylinder than the kinetic energy of the sphere? (Assume the toys roll without slipping.)

Answer 1

Given: Small Sphere toy: Mass of 0.025 kg and radius of R= 0.022 m,

Small cylinder toy: Mass of 0.025 kg and radius of , r= 0.013 m.

Transational speed, v = 0.42 m/s

The moment of inertia of a uniform solid sphere of Mass M and radius R is = (2/5)MR^2

The moment of inertia of a uniform solid cylinder of mass M and radius r is = (1/2)Mr^2

The no slip condition means that ?=??,

ν= ωR => ω=v/R ......(1)

ω= (0.42m/s)/(0.022m) = 19.09 s^-1

Expression for rotational kinetic energy of the toy in sphere shape,

Kr,sphere = (1/2)*Iω^2 .....(2)

Here I is the rotational inertia of sphere, I = (2/5)*M*R^2 .......(3)

Substitute the equation 1 and 3 in (2) equation,

Expression for rotational kinetic energy of the toy in cylindrical shape,

Kr,cylinder = (1/2)*Iω^2

The moment of inertia of a uniform solid cylinder of mass M and radius r is, I = (1/2)M*r^2

Kcylinder - Ksphere = (3.84*10^-4) - (8.81*10^-4) = -4.96 *10^-4 J