Question

a vertical wheel with a diameter of 50 cm starts from rest and rotates with a constant angular acceleration of 7.5 rad/s2 around a fixed axis through its center counterclockwise. where is the point that is initially at the bottom of the wheel at s? express your answer as an angle in radians between 0 and , relative to the positive axis. 4.7123 incorrect radians what is the point's tangential acceleration at this instant?

Answer 1

The point that is initially at the bottom of the wheel is located at an angle of 4.7123 radians, relative to the positive axis.

Step-by-step explanation:

The point that is initially at the bottom of the wheel is located at an angle of 4.7123 radians, relative to the positive axis. To find this, we need to calculate the angular displacement of the wheel. The formula for angular displacement is:

θ = (1/2) * α * t^2

where θ = angular displacement, α = angular acceleration, and t = time. Plugging in the given values:

θ = (1/2) * 7.5 * (10)^2 = 375 radians

Since the wheel is rotating counterclockwise, we add this angular displacement to the initial angle of 0 radians to get the final angle:

Final angle = 0 + 375 = 375 radians

However, the angle is given relative to the positive axis, so we subtract it from 2π radians (one complete revolution) to get:

Final angle = 2π - 375 = 4.7123 radians

Answer 2

The point that is initially at the bottom of the wheel at time s is 8.1687 radians and the tangential acceleration of the point at this instant is 1.875 m/s^2.

Step-by-step explanation:

To find the point that is initially at the bottom of the wheel at time s, we need to first calculate the angular displacement at that time. We can use the formula: π = θ + αt^2/2, where π is the angular displacement, θ is the initial angular position (0 radians in this case), α is the angular acceleration (7.5 rad/s^2), and t is the time. Rearranging the formula, we have: αt^2/2 = π - θ, which simplifies to: αt^2/2 = π. Solving for t, we get: t = sqrt(2π/α). Plugging in the values, we get: t = sqrt(2*3.1416/7.5) = 1.4702 seconds. Now, we can find the angular displacement (s) using the formula: s = θ + αt^2/2. Plugging in the values, we get: s = 0 + 7.5*(1.4702)^2/2 = 8.1687 radians.

The tangential acceleration of the point at this instant can be found using the formula: at = rα, where r is the radius of the wheel (diameter/2 = 50/2 = 25 cm = 0.25 m) and α is the angular acceleration (7.5 rad/s^2). Plugging in the values, we get: at = 0.25 * 7.5 = 1.875 m/s^2.