Rotating 2m Wheel: Angular Speed, Tangential Speed, and Total Acceleration

[jth23]

1. A wheel 2.00m in diameter lies in a vertical plane and rotates with a constant angular acceleration of 4.00 rad/s^2. The wheel starts at rest at t=0, and the radius vector of a certain point P on the rim makes an angle of 57.3 with the horizontal at this time. At t=2, find the angular speed of the wheel, the tangential speed and the total acceleration of the point P, and the angular position of the point P.



2.



3. for the speed I times the acceleration with the time to get 8 rad/s. The rest i don't understand so whatever help you could give will be grateful. Thanks.

 

[jbsteele]

Angular Speed: 8 rad/sTangential Speed: 12.57 m/sTotal Acceleration of Point P: 8.00 m/s^2Angular Position of Point P: 115.2 rad

 

[Moetasim]

I can provide a detailed response to your question. Let's break down the information given to us step by step.

1. The wheel has a diameter of 2.00m and is rotating in a vertical plane with a constant angular acceleration of 4.00 rad/s^2. This means that the wheel is increasing its angular velocity by 4.00 radians per second every second.

2. At t=0, the wheel is at rest and the radius vector of a certain point P on the rim makes an angle of 57.3 degrees (or 1 radian) with the horizontal. This means that the point P is located at the top of the wheel when it starts rotating.

3. At t=2, we are asked to find the angular speed of the wheel, the tangential speed and total acceleration of the point P, and the angular position of the point P.

To find the angular speed of the wheel, we can use the formula ω = ω0 + αt, where ω0 is the initial angular velocity (which is 0 since the wheel starts at rest), α is the angular acceleration (4.00 rad/s^2), and t is the time (2 seconds). Plugging in these values, we get:

ω = 0 + (4.00 rad/s^2)(2s) = 8.00 rad/s

This means that the wheel is rotating at a speed of 8.00 radians per second at t=2.

To find the tangential speed of the point P, we can use the formula v = rω, where v is the tangential speed, r is the radius of the wheel (1.00m), and ω is the angular speed we just calculated (8.00 rad/s). Plugging in these values, we get:

v = (1.00m)(8.00 rad/s) = 8.00 m/s

This means that the point P is moving at a speed of 8.00 meters per second at t=2.

To find the total acceleration of the point P, we can use the formula a = αr, where a is the total acceleration, α is the angular acceleration (4.00 rad/s^2), and r is the radius of the wheel (1.00m). Plugging in these values, we get:

a = (4.00 rad/s^2