Question

At t=300 s, a point on the rim of a 0.200 m radius wheel has a tangential speed of 50.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.0 m/s Calculate the wheel's constant angular acceleration b. Calculate the angular velocities at t03.00s at t=0 c. Through what angle did the wheel turn between t=0 and t=3.00 s ? d. At what time did the radial acceleration equal to g ?

Answer 1

To determine the wheel's angular velocity, we apply the formula ω = ω_0 + αt, where ω_0 is the initial angular velocity and α is the angular acceleration. Then to find the angle turned by the wheel, we use θ = ω_0t + ½αt^2. The tangential acceleration is r * α, linking angular and linear dynamics in rotational motion.

Step-by-step explanation:

To solve this problem in rotational dynamics, we use the relationships between angular velocity, angular acceleration, and the rotational variables that describe circular motion. Specifically, for part a, we will use the equation:

ω = ω0 + αt

For part b, we will use the formula for rotational displacement given a constant angular acceleration:

θ = ω0t + ½αt2

For part c, we find the tangential acceleration by using the relationship between angular and tangential acceleration:

at = r * α

The wheel has a tangential speed of 50.0 m/s at t=300s and a constant tangential acceleration of -10.0 m/s2 (negative since it's slowing down). To calculate the constant angular acceleration (α), we can use the relation at = r * α, where r is the radius of the wheel.

Also, using v = r * ω, we can calculate the angular velocity of the wheel (ω). For the angle turned between t=0s and t=3.00s, we can use the kinematic equations for rotational motion, as described in part b.

Answer 2

a. The wheels constant angular acceleration is
`\(\alpha = -50.0 \, \text{rad/s}^2\)`

b. The angular velocities at t = 3.00s is 250rad/s and at t=0 is 100 rad/s.

c. The angle turned by the wheel between
`\(t = 0\) and \(t = 3.00 \, \text{s}\)` is 75 rad/s.

d. Given
`\(\alpha = -50.0 \, \text{rad/s}^2\)`, it seems that the radial acceleration will never equal
`\(g\)` because the magnitude of
`\(\alpha\)` is less than
`\(g/r\)`.

Given:

• Radius of the wheel,
  `\(r = 0.200 \, \text{m}\)`
• Tangential speed at
  `\(t = 3.00 \, \text{s}\), \(v = 50.0 \, \text{m/s}\)`
• Tangential acceleration,
  `\(a_t = -10.0 \, \text{m/s}^2\)` (negative since the wheel is slowing down)

(a) Calculate the wheel's constant angular acceleration.

The tangential acceleration
`\(a_t\)` is related to the angular acceleration
`(\(\alpha\))` by the formula:
`\(a_t = r * \alpha\)`

Given
`\(a_t = -10.0 \, \text{m/s}^2\) and \(r = 0.200 \, \text{m}\)`:

`\(\alpha = (a_t)/(r)\)`

Let's calculate
`\(\alpha\)`:

`\(\alpha = \frac{-10.0 \, \text{m/s}^2}{0.200 \, \text{m}}\)`

`\(\alpha = -50.0 \, \text{rad/s}^2\)`

(b) Calculate the angular velocities at
`\(t = 3.00 \, \text{s}\) and \(t = 0\)`.

Given:

`\(v = 50.0 \, \text{m/s}\) \\\\r = 0.200 \, \text{m}\) \\\\\alpha = -50.0 \, \text{rad/s}^2\)`

Angular Velocity at
`\(t = 3.00 \, \text{s}\)`:

`\(\omega_(3s) = (v)/(r)\)\ \\\ \omega_(3s) = \frac{50.0 \, \text{m/s}}{0.200 \, \text{m}}\)\ \\\(\omega_(3s) = 250 \, \text{rad/s}\)`

Angular Velocity at
`\(t = 0\)`:

Using the kinematic equation:

`\(\omega_(0s) = \omega_(3s) + \alpha * t\)\(\omega_(0s) \\\ = 250 \, \text{rad/s} + (-50.0 \, \text{rad/s}^2) * 3.00 \, \text{s}\)\(\omega_(0s) \\\ = 250 \, \text{rad/s} - 150 \, \text{rad/s}\)\(\omega_(0s) \\\ = 100 \, \text{rad/s}\)`

(c) Calculate the angle turned by the wheel between
`\(t = 0\) and \(t = 3.00 \, \text{s}\)`.

The angular displacement (\(\theta\)) can be calculated using the formula:

`\(\theta = \omega_(0s) * t + (1)/(2) * \alpha * t^2\)`

Substituting the values:

`\(\theta = 100 \, \text{rad/s} * 3.00 \, \text{s} + (1)/(2) * (-50.0 \, \text{rad/s}^2) * (3.00 \, \text{s})^2\)\(\theta \\\\ \theta= 300 \, \text{rad} - 225 \, \text{rad}\)\(\theta \\\\ \theta= 75 \, \text{rad}\)`

(d) At what time will the radial acceleration equal
`\(g\)`?

The radial acceleration is given by
`\(a_r = r * \alpha\)`. Setting
`\(a_r\)` equal to
`\(g\)` (acceleration due to gravity):

`\(a_r = g = 9.81 \, \text{m/s}^2\)\(a_r = r * \alpha\)\(9.81 \, \text{m/s}^2 = 0.200 \, \text{m} * \alpha\)\(\alpha = \frac{9.81 \, \text{m/s}^2}{0.200 \, \text{m}}\)\(\alpha = 49.05 \, \text{rad/s}^2\)`

Given
`\(\alpha = -50.0 \, \text{rad/s}^2\)`, it seems that the radial acceleration will never equal
`\(g\)` because the magnitude of
`\(\alpha\)` is less than
`\(g/r\)`.