Question

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.

Answer 1

A) τ = 300 N*m

B) θ = 0.4 rad = 22.9°

Step-by-step 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°