The wheel having a mass of 100 kg and a radius of gyration about the z axis of kz=300mm, rests on the smooth horizontal plane. a. if the belt is subjected to a force of p=200n, determine the angular velocity of the wheel, three seconds after the force is applied. b. determine the speed of its center of mass o, three seconds after the force is applied.

a) 20 rad/s

b) 6 m/s

Explanation:

b) Force acting on the wheel is 200 N

mass of the wheel is 100 kg

From Newton's second law of motion, F = m × a

Where F is the net force acting on the body

m is mass of the body

a is the acceleration of the body

By substituting the values we get, a = 2 m/s²

As acceleration is constant, we can use the below formula for calculating the final velocity of the object

v = u + a × t

Where v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

u = 0 (∵ it starts from rest)

By substituting the values we get

v = 0 + 2 × 3 = 6 m/s

∴ Speed of center of mass after 3 seconds = 6 m/s

a) As the wheel rotates about z-axis, radius of gyration will be the radius of wheel

∴ Radius of the wheel = 300 mm

Torque acting on the wheel about axis of rotation = 300 mm × 200 N =

60 N·m

Torque = (Moment of inertia) × (angular acceleration)

Assuming that the mass of spokes of the wheel to be negligible,

Moment of inertia of the wheel about axis of rotation = 100 × 300² × = 9 kg·m²

Then,

60 = 9 × (angular acceleration)

∴ angular acceleration ≈ 6·67 rad/s²

As angular acceleration of the wheel is constant, we can use the below formula for calculation of final angular speed

= + α × t

Where

is the final angular velocity

is the initial angular velocity

α is the angular acceleration

t is the time taken

is 0 (∵ initially it starts from rest)

By substituting the values we get

= 6·67 × 3 = 20 rad/s

∴ Angular velocity of the wheel after three seconds = 20 rad/s

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures