A wheel rotating initially at an angular velocity of 1.0 rad/s, accelerates over a 10 S interval at a rate of 0.1 rad/s^2. What angular displacement does the wheel undergo in this 10 S interval, in terms of revolutions?

Why did the wheel go to therapy? Because it had trouble dealing with all the rotations!

But let's get to your question. To find the angular displacement of the wheel, we need to use the formula:

θ = ω0 * t + (1/2) * α * t^2

Where:
θ is the angular displacement,
ω0 is the initial angular velocity,
t is the time interval, and
α is the angular acceleration.

Plugging in the values:
ω0 = 1.0 rad/s
t = 10 s
α = 0.1 rad/s^2

θ = 1.0 rad/s * 10 s + (1/2) * 0.1 rad/s^2 * (10 s)^2

Calculating this equation will give us the angular displacement in radians. But you mentioned in terms of revolutions. Let me do the conversion for you.

1 revolution = 2π radians

So, the angular displacement in terms of revolutions is:

θ (rev) = θ (rad) / (2π)

Now, let me pull out my math hat and calculate the result for you... drum roll, please!

θ (rad) = 1.0 rad/s * 10 s + (1/2) * 0.1 rad/s^2 * (10 s)^2
θ (rad) = 10 rad + 1 rad = 11 rad

θ (rev) = 11 rad / (2π) ≈ 1.75 rev

Therefore, the wheel undergoes an angular displacement of approximately 1.75 revolutions in the 10 second interval. That's quite a spin!

To determine the angular displacement of the wheel in terms of revolutions, we can use the following formula:

Angular displacement (θ) = Initial angular velocity (ω) × time (t) + 0.5 × angular acceleration (α) × time² (t²)

Given:
Initial angular velocity (ω) = 1.0 rad/s
Angular acceleration (α) = 0.1 rad/s²
Time (t) = 10 s

Plugging in the values into the formula:

θ = (1.0 rad/s) × 10 s + 0.5 × (0.1 rad/s²) × (10 s)²

θ = 10 rad + 0.5 × 0.1 rad/s² × 100 s²

θ = 10 rad + 0.05 rad/s² × 10000 s²

θ = 10 rad + 500 rad

θ = 510 rad

To convert radians to revolutions, we need to divide the angular displacement by 2π.
Since 1 revolution is equal to 2π radians, we can use the conversion factor:

1 revolution = 2π radians

θ_revolutions = θ / (2π)

θ_revolutions = 510 rad / (2π)

θ_revolutions ≈ 81.01 revolutions (rounded to two decimal places)

Therefore, the wheel undergoes an angular displacement of approximately 81.01 revolutions in the 10-second interval.

To find the angular displacement, you need to use the kinematic equation for rotational motion, which relates angular displacement, initial angular velocity, angular acceleration, and time. The equation is:

θ = ω₀t + (1/2)αt²

where:
θ is the angular displacement (in radians)
ω₀ is the initial angular velocity (in rad/s)
α is the angular acceleration (in rad/s²)
t is the time interval (in seconds)

Given:
ω₀ = 1.0 rad/s (initial angular velocity)
α = 0.1 rad/s² (angular acceleration)
t = 10 S (time interval)

Substituting the given values into the equation:

θ = (1.0 rad/s)(10 s) + (1/2)(0.1 rad/s²)(10 s)²

Simplifying:

θ = 10 rad + 0.5 rad = 10.5 rad

To convert the angular displacement from radians to revolutions, you need to know that 1 revolution is equal to 2π radians. Therefore, to find the angular displacement in terms of revolutions:

θ (in revolutions) = θ (in radians) / (2π)

θ (in revolutions) = 10.5 rad / (2π)

θ (in revolutions) ≈ 1.67 revolutions

Therefore, the wheel undergoes an angular displacement of approximately 1.67 revolutions in the 10-second interval.

D = Vo*t + 0.5a*t^2.

D = 1*10 + 0.05*10^2 = 15 rad.

D = 15rad/(6.28rad/rev.) = 2.39 rev.