Circular Motion Physics Example 4

Follow the full solution, then compare it with the other examples linked below.

Example 4

medium
A point on the rim of a bicycle wheel (r=0.35ย mr = 0.35 \text{ m}) moves at 5ย m/s5 \text{ m/s}. What is the period of rotation and the centripetal acceleration?

Solution

  1. 1
    Period: T=2ฯ€rv=2ฯ€ร—0.355=2.205=0.44ย sT = \frac{2\pi r}{v} = \frac{2\pi \times 0.35}{5} = \frac{2.20}{5} = 0.44 \text{ s}.
  2. 2
    Centripetal acceleration: ac=v2r=250.35โ‰ˆ71.4ย m/s2a_c = \frac{v^2}{r} = \frac{25}{0.35} \approx 71.4 \text{ m/s}^2

Answer

Tโ‰ˆ0.44ย s,acโ‰ˆ71.4ย m/s2T \approx 0.44 \text{ s}, \quad a_c \approx 71.4 \text{ m/s}^2
Points on a spinning wheel undergo circular motion. The smaller the radius at a given speed, the larger the centripetal acceleration and the shorter the period of rotation.

About Circular Motion

Motion of an object along a circular path where the speed may be constant but the velocity is continuously changing direction, requiring a centripetal acceleration.

Learn more about Circular Motion โ†’

More Circular Motion Examples

Example 1 easy

A car drives around a circular track of radius [formula] at a constant speed of [formula]. What is t

Example 2 medium

A satellite orbits Earth at a radius of [formula] with a period of [formula]. What is its orbital sp

Example 3 medium

A car travels around a circular track of radius 50 m at 20 m/s. Find the centripetal acceleration.

Example 5 hard

A ball on a [formula] string is swung in a vertical circle. What minimum speed must it have at the t

โ† All Circular Motion Examples Circular Motion Concept