Question

to answer this question, first click here to view the associated video. in which of the following is a tangential acceleration involved in the motion? a runner running at a constant 4 m/s around a circular track a 1 - kg mass swinging freely at the end of a rope the outer edge of merry - go - round ride rotating at a constant 30°/s the tip of a lawn mower blade running at constant 40 rev/s

Explanation:

Tangential acceleration \(a_t = r\alpha\), where \(\alpha=\frac{d\omega}{dt}\) (angular acceleration). For tangential acceleration to exist, the angular velocity \(\omega\) must be changing (non - constant).

• A runner running at a constant \(4\ m/s\) around a circular track: The speed (magnitude of velocity) is constant. The acceleration is centripetal (\(a_c=\frac{v^{2}}{r}\)), no tangential acceleration (\(a_t = 0\) since \(\frac{dv}{dt}=0\)).
• A \(1 - kg\) mass swinging freely at the end of a rope: As the mass swings, its speed changes. The angular velocity \(\omega\) changes. Since \(a_t=r\alpha\) and \(\alpha=\frac{d\omega}{dt}

eq0\) (because the mass is speeding up and slowing down during the swing), there is a tangential acceleration.

• The outer edge of a merry - go - round ride rotating at a constant \(30^{\circ}/s\): The angular velocity \(\omega\) is constant (\(\frac{d\omega}{dt} = 0\)). So, \(a_t=r\alpha=r\frac{d\omega}{dt}=0\).
• The tip of a lawn mower blade running at a constant \(40\ rev/s\): The angular velocity \(\omega\) is constant (\(\frac{d\omega}{dt}=0\)). So, \(a_t=r\alpha = r\frac{d\omega}{dt}=0\).

Answer:

A \(1 - kg\) mass swinging freely at the end of a rope.