Amerry-go-round on a playground consists of a horizontal solid disk with a weight of 810 n and a radius of 1.51 m. a child applies a force 51.0 n tangentially to the edge of the disk to start it from rest. what is the kinetic energy of the merry-go-round disk (in j) after 2.95 s?

K = 113.55 J

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)

Calculation of the torque (τ) applied to the merry-go-round

τ = F * R Formula  (2)

Where :

F is the tangential force in Newtons (N)

R is the radio , perpendicular to F

Data

F=51.0 N

R =  1.51 m

We replace data in the formula (2)

τ = F * R

τ = 51.0 N * 1.51 m = 77.01 N*m

Moment applied to the merry-go-round:

The moment of inertia of a solid disk that rotates around an axis that passes through its center is::

I = (1/2)m*R²  Formula (3)

Where:

I: moment of inertia of the solid disk ( kg*m²)

m : mass (kg)

R : disk radio

Data

W=810 N :weight of the merry-go-round

W= m*g

g= 9.8 m/s² = acceleration due to gravity

m = W/g =810/ 9.8

m = 82.65 kg : mass of the merry-go-round

R =  1.51 m : radio of the merry-go-round

We replace data in the formula (3)

I =  (1/2)(82.65)*(1.51)²

I = 94.23 kg*m²

Angular acceleration of the  merry-go-round

We replace data in the formula (1):

τ = I * α

77.01= 94.23 * α

α= 77.01 / 94.23

α = 0.8173 rad/s²

Kinetic energy of the merry-go-round disk

K = (1/2) I *ω² Formula (4)

where:  

K : rotation kinetic energy (J)

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

ω: Angular speed (rad/s)  

Kinematics of the merry-go-round

We apply the equations of circular motion uniformly accelerated:

ω²= ω₀+α*t  Formula (5)

Where:  

α : Angular acceleration (rad/s²)  

ω₀ : Initial angular speed ( rad/s)  

ω : Final angular speed ( rad/s)

t :  time interval (s)

Data  

α = 0.8173 rad/s²

ω₀ = 0

t =  2.95 s

We replace data in Formula (5)

ω²= ω₀+α*t

ω²= 0+ (0.8173) *(2.95)

ω²= 2.41

ω = 1.55 rad/s

We replace I = 94.23 kg*m² and ω = 1.55 rad/s in the formula (4):

K = (1/2) (94.23)*(1.55)²

K = 113.55 J

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student

a.   mass number from atomic number