Question

A certain wheel is a uniform disk of radius R = 0.5 m and mass M = 10.0 kg. A constant force of Fapp = 15.0 N is applied to the center of mass of the wheel in the positive x-direction. The wheel rolls along the ground without slipping.

(a) Compute the rotational inertia of the wheel about its center of mass.

(b) Compute the magnitude and direction of the friction force acting on the wheel from the ground.

(c) If the wheel is at rest at t = 0s, what is its kinetic energy at t = 5.0s?

(d) How much time is required for the wheel acquire a kinetic energy of 1500 J?

Answer 1

a)

Rotational Inertia of wheel which is a uniform disc about center of mass, I = 1/2 * M*R²

= 1/2 *10 * 0.5² = 1.25 kgm²

b)

Let a be the acceleration of center of mass of the wheel,

F is the force applied in the x-direction

consider point A which touching the ground, when the wheel will move in +-ve x-direction, point A has tendency to move back, so to oppose the motion of point A , a frictional force will act in positive x-direction as shown in figure,

Apply Newton's second law in x-direction,

F + f = M*a

15 + f = 10*a ....(1)

for puré rolling motion,

torque, T = i*

só, (F-f) * R = I * a/R (since angular acceleration, = a/R)

(15-f) * 0.5 = 1.25 *a/0.5

15 - f = 5a

30 - 2f = 10a ...(2)

subtracting (1) from (2)

30 -2f - (15 + f) = 0

15 = 3f

so, frictional force, f = 5N in positive x-direction

c)

so, 15 + 5 = 10 a

so, acceleration, a = 2 m/s²

apply. first law of motion, v = u + at

at t= 0. u = 0

so, V₅ = 2*5 = 10 m/s

Total kinetic energy of the wheel, = 1/2 * I *w² + 1/2 *M*V²

= 1/2 * 1/2 * M*R²*(V/R)² + 1/2 * M *V²

  = 1/4 * M*V² + 1/2 * MV² = 3/4 * MV²

so, Kinetic energy at t= 5 second, KE₅ = 3/4 * 10 *10² = 750 J

d)

KE = 3/4 * MV²

1500 = 3/4 * 10 * V²

so, V = 14.142 m/s

as V = u + at

14.142 = 0 + 2*t

só. time,t = 7.071 second