Question

A wheel of radius 0.38 m rotates in a clockwise sense about a fixed axle with negligible friction at an initial angular speed of 1.6 rad/s. The wheel’s mass of 14 kg is concentrated in its rim. You apply a clockwise torque to the rotating wheel by pushing on the rim tangentially with a constant force of 24 N.
Find the wheel's angular speed, in radians per second, 0.13 seconds after you start pushing on the rim.

Answer 1

Answer:

ωf = 2.19 rad/s

Explanation:

Newton's second law:

F = ma has the equivalent for rotation:

τ = I * α   Formula  (1)

where:

τ : It is the torque applied to the body.  (Nxm)

I :  it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Data

F = 24 N  : Tangential force

m : 14 kg : mass of the wheel

R = 0.38 m : radius of the wheel

Moment  of inertia of  the wheel

The moment of inertia of wheel is defined as follows:

I = m* R²

I = 14 kg*(0.38)²

I = 2.0216 kg*m²

Torque applied to the wheel

The Torque ( τ) applied to the wheel is defined as follows:

τ = F*d

Where:

F : Tangential force applied to the wheel

d : Perpendicular distance of the tangential force to the axis of rotation

τ = (24N)*(0.38 m)

τ = 9.12 N*m

Angular acceleration of the wheel  (α )

We replace data in the formula (1):

τ = I * α

9.12 = (2.0216) * α

α= 9.12 / (2.0216)

α = 4.5 rad/s²  

Kinematics of the wheel

We apply the equations of circular motion uniformly accelerated :  

ωf = ω₀ + α*t Formula (2)

Where:  

α : Angular acceleration (rad/s²)  

ω₀ : Initial angular speed ( rad/s)  

ωf : Final angular speed ( rad

t : time interval (s)

Data  

α = 4.5 rad/s²  

ω₀ =1.6 rad/s

t =  0.13 s    

We replace data in the formula (2):  

ωf = ω₀ + α*t

ωf = 1.6 + (4.5)*(0.13)

ωf = 1.6 + (4.5)*(0.13)

ωf = 2.19 rad/s