How Do You Calculate the Angular Position of a Point on a Rotating Wheel?

[bcd201115]

Homework Statement

a wheel 2m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4rad/s^2. The wheel starts 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=2s find d.) the angular position for point P.

Homework Equations

The answer for d.) is 9radians but can someone help me figure out how to get this. Also if you could help show me how to get the direction of the total acceleration.

The Attempt at a Solution

I already found that angular speed, ω=8rad/s, tangential speed v=8m/s, and total acceleration is 64.12m/s^2

 

[collinsmark]

Homework Statement

a wheel 2m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4rad/s^2. The wheel starts 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=2s find d.) the angular position for point P.

Homework Equations

The answer for d.) is 9radians but can someone help me figure out how to get this.

All of your kinematics equations for uniform (linear) acceleration have corresponding versions for uniform angular acceleration. Just replace s with \theta, v with \omega, and a with \alpha.

For example one of your linear kinematics equations is:

s = s_0 + v_0t + \frac{1}{2}at^2.

The corresponding angular version is simply

\theta = \theta_0 + \omega_0t + \frac{1}{2}\alpha t^2.

Also if you could help show me how to get the direction of the total acceleration.

Just use trigonometry. The centripetal part of the linear acceleration is in the radial direction. The tangential part in the tangential direction. These directions are perpendicular to one another. So the inverse tangent applies. (Draw the acceleration vectors on a piece of paper. Each of these directions are components of the resulting linear acceleration vector, and they form a triangle.)

The Attempt at a Solution

I already found that angular speed, ω=8rad/s,

Well, the wheel is accelerating (rotationally), so it doesn't have a constant angular speed. But yes, 8 rad/sec is the final angular speed at t = 2 sec.

tangential speed v=8m/s,

Okay, that's the final tangential (linear) speed at t = 2 sec.

and total acceleration is 64.12m/s^2

That looks right to me, for the magnitude.

 

[bcd201115]

in the equation θ_f=θ_i+ω_it+(1/2)αt² what would i put for θ_f and θ_i?

 

[collinsmark]

in the equation θ_f=θ_i+ω_it+(1/2)αt² what would i put for θ_f and θ_i?

θ_f is what you're trying to find out: the angular position for point P, at time t = 2 sec.

θ_i is the initial angular position (the angular position, in radians, at time t = 0). (Hint: You'll have to convert 57.3^o to radians.)