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.)

 

[asteriatic]

.

I would approach this problem by first clarifying the given information and understanding the concepts involved. The problem states that a wheel with a diameter of 2m is rotating about its central axis with a constant angular acceleration of 4rad/s^2. This means that the angular velocity (ω) of the wheel is increasing at a rate of 4rad/s every second. We also know that the wheel starts at t=0, and at this time, the radius vector of point P on the rim makes an angle of 57.3° with the horizontal.

To find the angular position (θ) of point P at t=2s, we can use the equation θ=θ0+ω0t+1/2αt^2, where θ0 is the initial angular position, ω0 is the initial angular velocity, α is the angular acceleration, and t is the time. Since we are given the initial angle (57.3°), we can convert it to radians (57.3°=π/180 radians). Plugging in the values, we get θ=π/180+8(2)+1/2(4)(2^2)=π/180+16+8=16.056 radians. This is the angular position of point P at t=2s.

To determine the direction of the total acceleration, we can use the equation a=αr, where a is the total acceleration, α is the angular acceleration, and r is the radius of the wheel (1m in this case, since the diameter is 2m). Plugging in the values, we get a=4(1)=4m/s^2. Since the total acceleration is in the same direction as the angular acceleration, the direction of the total acceleration is also counterclockwise.

In conclusion, the angular position of point P at t=2s is 16.056 radians and the direction of the total acceleration is counterclockwise.