Question

1. a bicycle is travelling at a speed of 5.00 m/s. if the radius of the wheel is 32.0 cm, then its frequency of rotation is ______ hz. (2 marks)

r = 32.0 → 0.320 m
ω = v / r = 5.00 / 0.320 = 15.625 rad/s
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1. a stunt rider on a motorcycle rides down a ramp and into a vertical loop - the - loop. if the diameter of the loop is 7.50 m, then the slowest speed the motorcycle can have at the top of the loop if it is to remain in contact with the loop is ______ m/s. (2 marks)

Explanation:

Step1: Convert radius to meters

The radius \( r = 32.0\space cm = 0.320\space m \).

Step2: Find the circumference of the wheel

The circumference \( C = 2\pi r \). Substituting \( r = 0.320\space m \), we get \( C = 2\pi\times0.320 \approx 2.0106\space m \).

Step3: Relate speed, circumference, and frequency

The speed \( v \) of the bicycle is related to the frequency \( f \) and circumference \( C \) by \( v = f\times C \). So, \( f=\frac{v}{C} \).
Substituting \( v = 5.00\space m/s \) and \( C\approx2.0106\space m \), we get \( f=\frac{5.00}{2.0106}\approx2.49\space Hz \).

For the second problem:

Step1: Identify the condition for contact at the top

At the top of the loop, the minimum speed occurs when the normal force is zero, and gravity provides the centripetal force. So, \( mg=\frac{mv^{2}}{r} \), where \( r \) is the radius of the loop.

Step2: Find the radius of the loop

The diameter \( d = 7.90\space m \), so radius \( r=\frac{d}{2}=\frac{7.90}{2}=3.95\space m \).

Step3: Solve for speed

From \( mg=\frac{mv^{2}}{r} \), we can cancel \( m \) and solve for \( v \): \( v = \sqrt{gr} \). Substituting \( g = 9.8\space m/s^{2} \) and \( r = 3.95\space m \), we get \( v=\sqrt{9.8\times3.95}=\sqrt{38.71}\approx6.22\space m/s \).

Answer:

1. \(\approx2.49\space Hz\)
2. \(\approx6.22\space m/s\)