Question

A uniform disc of radius 30 cm and mass 12 kg is pivoted so that it is free to rotate about an axis through its center. A string wrapped around the disc is pulled with a force of 80 N. What is the angular displacement of the disc in the first 5 s ?

Answer 1

θ= 555.55 rad

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 : moment of inertia: 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)

Moment of inertia of the disc that rotates around an axis through its center

I= (1/2) m*R² Formula (2)

Where:

m:mass of the disc (kg)

R: radius of the disc (m)

Torque applied to disc (τ)

τ = F*R Formula (3)

Where

F: tangential force applied to the disc

R :Distance from F to the axis of rotation (m)

Data

m : 12 kg

R = 30 cm= 0.3 m

F= 80 N

Problem Development

We replace Formula (2) and Formula (3) in the formula (1)

τ = I * α

F*R= (1/2) m*R²* α

α =
`\alpha = (F*R)/((1)/(2)*m*R^(2)  )`

`\alpha =(2F)/(m*R)`

`\alpha =(2*80)/(12*0.3)`

α = 44.4 rad/s²

Kinematics of the disc

We apply the equations of circular motion uniformly accelerated

θ= ω₀ t+ (1/2)α*t² Formula (4)

Where:

α : Angular acceleration (rad/s²)

ω₀ : Initial angular speed ( rad/s)

t: time interval (s)

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

Data

ω₀= 0

α = 44.4 rad/s²

t =5 s

We replace data in the formula (4)

θ= 0+ (1/2)(44.4)*(5)²

θ= 555.55 rad