Question

The angular velocity of a flywheel of radius 1.7m varies as Ω(t)=1.15t (rad/s). Plot the magnitudes of the tangential and of the centripetal accelerations of a point located on the outer edge of this flywheel as a function of time for t∈(0,3)s?

Answer 1

For a flywheel of 1.7 m radius with angular velocity Ω(t) = 1.15t, the tangential acceleration (at) is constant at 1.955 m/s², and the centripetal acceleration (ac) increases with time, being proportional to the square of time, t².

Step-by-step explanation:

When a flywheel of radius 1.7 m has an angular velocity that varies according to Ω(t) = 1.15t, we can calculate both the tangential and centripetal accelerations of a point on its edge. For the tangential acceleration, at, which is the rate of change of tangential speed, it is given by the product of the radius and the angular acceleration (which is the derivative of angular velocity Ω(t)).

Ω(t) = 1.15t implies an angular acceleration, α = dΩ/dt = 1.15 rad/s². Therefore, at = rα = 1.7 m × 1.15 rad/s² = 1.955 m/s². This tangent acceleration is constant over time since α is constant.

For the centripetal acceleration, ac, which depends on the square of the tangential speed divided by the radius, we need to first find the tangential speed (v = rΩ). Since Ω(t) = 1.15t, the tangential speed v(t) = rΩ(t) = 1.7 m × 1.15t m/s. Therefore, ac = v(t)²/r = (1.7 × 1.15t)² / 1.7 = 1.955t² m/s².

The magnitude of the tangential acceleration remains constant at 1.955 m/s², while the magnitude of the centripetal acceleration increases quadratically with time, proportional to the square of time, t².