8. a baseball batter angularly accelerates a bat from rest to 20 rad/s in 40 ms (milliseconds). if the bat’s moment of inertia is 0.6 kg m2, then finda) the torque applied to the bat andb) the angle through which the bat moved.

A) τ = 300 N*m

B) θ  = 0.4 rad = 22.9°

Explanation:

Newton's second law:

F = ma has the equivalent for rotation:

τ = I * α   Formula  (1)

where:

τ : It is the moment 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

I = 0.6 kg*m² : moment of inertia of the bat

Angular acceleration of the bat

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

ω₀ = 0

ωf =  20 rad/s

t = 40 ms = 0.04 s

We replace data in the formula (2) :

ωf= ω₀ + α*t

20 = 0 + α* (0.04)

α = 20/ (0.04)

α = 500 rad/s²

Newton's second law to the bat

τ = (0.6 kg*m²) *(500 rad/s²) = 300 (kg*m/s²)* m

τ = 300 N*m

B)  Angle through which the bat moved.

We apply the equations of circular motion uniformly accelerated :  

ωf²= ω₀ ²+ 2α*θ Formula (3)

Where:  

θ : Angle that the body has rotated in a given time interval (rad)

We replace data in the formula (3):  

ωf²= ω₀²+ 2α*θ

(20)²= (0)²+ 2(500 )*θ

400 = 1000*θ

θ  = 400/1000

θ  = 0.4 rad

π rad = 180°

θ  = 0.4 rad *(180°/π rad)

θ  = 22.9°

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