Question

a wheel with radius 0.501 m rotates 5.67 times every second. find the period t of this motion. t = s what is the tangential speed v of a wad of chewing gum stuck to the rim of the wheel? v = m/s

Explanation:

Part 1: Find the period \( T \)

Step 1: Recall the relationship between frequency and period

The frequency \( f \) is the number of rotations per second. Here, \( f = 5.67 \, \text{Hz} \) (since it rotates 5.67 times every second). The period \( T \) is the reciprocal of the frequency, so \( T=\frac{1}{f} \).

Step 2: Calculate the period

Substitute \( f = 5.67 \, \text{Hz} \) into the formula: \( T=\frac{1}{5.67} \approx 0.1764 \, \text{s} \)

Part 2: Find the tangential speed \( v \)

Step 1: Recall the formula for tangential speed

The tangential speed \( v \) of a point on a rotating object is given by \( v = \frac{2\pi r}{T} \), where \( r \) is the radius of the wheel and \( T \) is the period. We can also use \( v = 2\pi r f \) (since \( T=\frac{1}{f} \), so substituting gives \( v = 2\pi r f \)).

Step 2: Substitute the values

We know \( r = 0.501 \, \text{m} \) and \( f = 5.67 \, \text{Hz} \). Plugging these into the formula \( v = 2\pi r f \):
\( v = 2\times\pi\times0.501\times5.67 \)
First, calculate \( 2\times0.501 = 1.002 \)
Then, \( 1.002\times5.67 \approx 5.68134 \)
Then, \( 5.68134\times\pi \approx 17.85 \, \text{m/s} \) (using \( \pi \approx 3.1416 \))

Answer:

s:
For the period \( T \): \( \approx 0.176 \, \text{s} \) (more precisely \( \frac{1}{5.67}\approx0.1764 \, \text{s} \))
For the tangential speed \( v \): \( \approx 17.9 \, \text{m/s} \) (more precisely \( \approx 17.85 \, \text{m/s} \))

(If we use more precise calculation for \( T=\frac{1}{5.67}\approx0.176367 \, \text{s} \), then \( v=\frac{2\pi\times0.501}{0.176367}\approx\frac{3.154}{0.176367}\approx17.88 \, \text{m/s} \))