Question

A wheel 2.00m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.00 rad/s². 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=2.00s , find(d) the angular position.

Answer 1

To find the angular position of the wheel at t = 2.00 s, we can use the equation θ = θ0 + (1/2)αt², where θ is the angular position, θ0 is the initial angular position, α is the angular acceleration, and t is the time. Given that the wheel starts at rest, the initial angular velocity is 0 rad/s. Plugging in the values, we find that the angular position at t = 2.00 s is 65.3°.

Step-by-step explanation:

To find the angular position of the wheel at t = 2.00 s, we need to use the equation:

θ = θ0 + ω0t + (1/2)αt²,

where θ is the angular position, θ0 is the initial angular position, ω0 is the initial angular velocity, α is the angular acceleration, and t is the time.

Given that the wheel starts at rest, the initial angular velocity is 0 rad/s. Therefore, the equation simplifies to:

θ = θ0 + (1/2)αt².

Using the given values, we can calculate the angular position at t = 2.00 s:

θ = 57.3° + (1/2)(4.00 rad/s²)(2.00 s)² = 57.3° + 8.00 rad = 65.3°.

Answer 2

The angular position of point P on the wheel at time t = 2.00 seconds, with an initial angle of 57.3° and an angular acceleration of 4.00 rad/s², is 17 radians.

Step-by-step explanation:

You want to find the angular position of a point on a wheel at a certain time given the initial angle, angular acceleration, and time that has elapsed. The initial angle is given as 57.3° (which is equivalent to 1 radian), the angular acceleration is 4.00 rad/s², and the time is 2.00 seconds. To calculate the angular position, θ, we can use the following kinematic equation for rotational motion:

θ = θ0 + ω0t + ½αt²

Where θ0 is the initial angular position, ω0 is the initial angular velocity (0 rad/s since the wheel starts at rest), α is the angular acceleration, and t is the time.

Let's plug in the known values:

• θ0 = 1 rad
• ω0 = 0 rad/s
• α = 4.00 rad/s²
• t = 2.00 s

Substituting these into the formula, we get:

θ = 1 rad + (0 rad/s × 2.00 s) + ½(4.00 rad/s²)(2.00 s)²

θ = 1 rad + 0 rad + (0.5 × 4.00 rad/s² × 4.00 s²)

θ = 1 rad + 16 rad

θ = 17 rad

The angular position of point P on the wheel at t = 2.00s is 17 radians.