Question

2 formula 1 point a student whirls a small box on a string above their head in a circular path. the radius of the path is equal to 0.7 meters. it takes 5.5 seconds for the box to complete 16 revolutions. what must be the tangential velocity of the box answer previous next

Explanation:

Step1: Find the period (time per revolution)

First, we need to find the time for one revolution (period \( T \)). The box completes 16 revolutions in 5.5 seconds, so \( T=\frac{5.5\ s}{16} \).

Step2: Calculate the circumference of the circular path

The circumference \( C \) of a circle is given by \( C = 2\pi r \), where \( r = 0.7\ m \). So \( C=2\pi(0.7)=1.4\pi\ m \).

Step3: Find the tangential velocity

Tangential velocity \( v \) is the distance traveled per unit time. The distance per revolution is the circumference, and the time per revolution is the period. So \( v=\frac{C}{T} \). Substituting \( C = 1.4\pi \) and \( T=\frac{5.5}{16} \), we get \( v=\frac{1.4\pi}{\frac{5.5}{16}}=\frac{1.4\pi\times16}{5.5} \). Calculating this: \( 1.4\times16 = 22.4 \), so \( v=\frac{22.4\pi}{5.5}\approx\frac{22.4\times3.1416}{5.5}\approx\frac{70.37184}{5.5}\approx12.79\ m/s \) (or more precisely, let's do the fraction: \( \frac{22.4}{5.5}=\frac{224}{55}\approx4.0727 \), so \( v = 4.0727\times\pi\approx12.8\ m/s \))

Answer:

Approximately \( 12.8\ m/s \) (or more precisely, using \( \pi\approx3.1416 \), \( v=\frac{1.4\times2\times\pi\times16}{5.5\times2}=\frac{44.8\pi}{11}\approx\frac{44.8\times3.1416}{11}\approx\frac{140.74368}{11}\approx12.79\ m/s \))