Question

is the same in each case. how does the net force in case a compare to the net force in case b?
case a
v, ½·m
2·r
the net force in case a is...
tap to select answer.
... the net force of case b.

Explanation:

Step1: Recall the formula for centripetal force

The formula for centripetal force is \(F = \frac{mv^{2}}{r}\).

Step2: Calculate the net force (centripetal force) for Case A

For Case A, \(m = \frac{1}{2}m_0\), \(r = 2R\), and \(v\) is the same as in Case B. Substituting into the formula: \(F_A=\frac{(\frac{1}{2}m_0)v^{2}}{2R}=\frac{m_0v^{2}}{4R}\).

Step3: Calculate the net force (centripetal force) for Case B

For Case B, \(m = m_0\), \(r = R\). Substituting into the formula: \(F_B=\frac{m_0v^{2}}{R}\).

Step4: Compare \(F_A\) and \(F_B\)

Divide \(F_A\) by \(F_B\): \(\frac{F_A}{F_B}=\frac{\frac{m_0v^{2}}{4R}}{\frac{m_0v^{2}}{R}}=\frac{1}{4}\). So \(F_A=\frac{1}{4}F_B\).

Answer:

The net force in Case A is one - fourth of the net force in Case B.