Question

A 42.0-cm diameter disk rotates with a constant angular acceleration of 2.70 rad/s2. It starts from rest at t = 0, and a line drawn from the center of the disk to a point P on the rim of the disk makes an angle of 57.3° with the positive x-axis at this time.

(a) Find the angular speed of the wheel at t=2.30s.

(b) Find the linear velocity and tangential acceleration of P at t=2.30s.

(c) Find the position of P (in degrees, with respect to the positive x-axis) at t = 2.30s.

Answer 1

The angular speed of the wheel at t=2.30s is 6.21 rad/s. The linear velocity of point P at t=2.30s is 1.30 m/s, and the tangential acceleration of P at t=2.30s is 0.57 m/s². The position of P at t=2.30s is 11.5 rad with respect to the positive x-axis.

Step-by-step explanation:

(a) To find the angular speed of the wheel at t = 2.30s, we can use the formula:

ω = ω0 + αt

Where:

ω = angular speed at time t

ω0 = initial angular speed

α = angular acceleration

t = time

Given:

ω0 = 0 rad/s (since the wheel starts from rest)

α = 2.70 rad/s2

t = 2.30s

Substituting the values into the equation:

ω = 0 + 2.70 * 2.30 = 6.21 rad/s

(b) To find the linear velocity of point P at t = 2.30s, we can use the formula:

v = rω

Where:

v = linear velocity

r = radius of the wheel (half of the diameter)

ω = angular speed at time t

Given:

r = 21.0 cm = 0.21 m

ω = 6.21 rad/s (from part a)

Substituting the values into the equation:

v = 0.21 * 6.21 = 1.30 m/s

To find the tangential acceleration of P at t = 2.30s, we can use the formula:

at = rα

Where:

at = tangential acceleration

r = radius of the wheel

α = angular acceleration

Given:

r = 21.0 cm = 0.21 m

α = 2.70 rad/s2

Substituting the values into the equation:

at = 0.21 * 2.70 = 0.57 m/s2

(c) To find the position of P at t = 2.30s, we can use the formula:

θ = θ0 + ω0t + 0.5αt2

Where:

θ = angular position

θ0 = initial angular position

ω0 = initial angular speed

α = angular acceleration

t = time

Given:

θ0 = 57.3° = 1 rad (since 1 rad = 57.3°)

ω0 = 0 rad/s (since the wheel starts from rest)

α = 2.70 rad/s2

t = 2.30s

Substituting the values into the equation:

θ = 1 + 0 + 0.5 * 2.70 * (2.30)2 = 11.5 rad

Answer 2

6.21 rad/s

1.3041 m/s, 0.567 m/s²

`106.4778\ ^(\circ)`

Step-by-step explanation:

`\omega_f` = Final angular velocity

`\omega_i` = Initial angular velocity = 0

`\alpha` = Angular acceleration = 2.3 rad/s²

`\theta` = Angle of rotation

t = Time taken = 2.3 s

Equation of rotational motion

`\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=0+2.7* 2.3\\\Rightarrow \omega_f=6.21\ rad/s`

The angular speed is 6.21 rad/s

Linear velocity is given by

`v=r\omega\\\Rightarrow v=0.21* 6.21\\\Rightarrow v=1.3041\ m/s`

Linear velocity is 1.3041 m/s

Tangential acceleration is given by

`a_t=r\alpha\\\Rightarrow a_t=0.21* 2.7\\\Rightarrow a_t=0.567\ m/s^2`

Tangential acceleration is 0.567 m/s²

`\theta=\omega_it+(1)/(2)\alpha t^2\\\Rightarrow \theta=0* t+(1)/(2)* 2.7* 2.3^2\\\Rightarrow \theta=7.1415\ rad`

In degress the angle would be

`57.3+7.1415* (180)/(\pi)=466.47780\ ^(\circ)`

From x axis it would be

`466.47780-360=106.4778\ ^(\circ)`

The angle is
`106.4778\ ^(\circ)` from x axis