Question

a 0.004 kg marble rolls horizontally around the inside edge of a hoop with a diameter of 40 cm at a constant speed of 2 m/s. determine the normal force of the wooden hoop against the marble.

Explanation:

Step1: Convert diameter to radius

The diameter \(d = 40\space cm=0.4\space m\), so the radius \(r=\frac{d}{2}=\frac{0.4}{2}= 0.2\space m\)

Step2: Use centripetal force formula

The normal force \(N\) provides the centripetal force \(F_c\). The formula for centripetal force is \(F_c=\frac{mv^{2}}{r}\), where \(m = 0.004\space kg\), \(v = 2\space m/s\) and \(r=0.2\space m\)
Substitute the values: \(N=\frac{0.004\times(2)^{2}}{0.2}\)

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Answer:

The normal force of the wooden hoop against the marble is \(0.08\space N\)