Problem Statement
A ball is threaded on a circular loop of radius $R$. What minimum speed must it have at the top to maintain contact?
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Circular motion requires a centripetal force directed toward the centre, providing the centripetal acceleration $a_c = v^2/r = \omega^2 r$. This force is not a new type of force — it is always the resultant of real forces (tension, normal force, friction, gravity) directed inward. At the minimum speed for maintaining contact, the normal force drops to zero.
- $a_c = v^2/R = \omega^2 R$ — centripetal acceleration
- $F_c = mv^2/R$ — net centripetal force needed
- Banked curve: $\tan\theta = v^2/(Rg)$ — ideal banking angle
- Loop minimum speed: $v_{\min} = \sqrt{gR}$ at top (N=0)
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: At the top, both gravity and the normal force point downward (toward center).
Step 2 — Apply the relevant physical law or equation: For minimum contact, $N = 0$:
Step 3 — Solve algebraically for the unknown: $$mg = \frac{mv_{top}^2}{R} \implies v_{top,\min} = \sqrt{gR}$$
Step 4 — Substitute numerical values with units: Minimum speed at bottom (energy conservation from bottom to top):
Step 5 — Compute and check the result: $$\frac12 mv_{bot}^2 = \frac12 mv_{top}^2 + mg(2R)$$
$$v_{bot,\min} = \sqrt{v_{top,\min}^2 + 4gR} = \sqrt{gR+4gR} = \sqrt{5gR}$$
Step 6: Tension at bottom with minimum speed:
Worked Calculation
$$mg = \frac{mv_{top}^2}{R} \implies v_{top,\min} = \sqrt{gR}$$
$$\frac12 mv_{bot}^2 = \frac12 mv_{top}^2 + mg(2R)$$
$$v_{bot,\min} = \sqrt{v_{top,\min}^2 + 4gR} = \sqrt{gR+4gR} = \sqrt{5gR}$$
At the top, both gravity and the normal force point downward (toward center).
For minimum contact, $N = 0$:
$$mg = \frac{mv_{top}^2}{R} \implies v_{top,\min} = \sqrt{gR}$$
Minimum speed at bottom (energy conservation from bottom to top):
$$\frac12 mv_{bot}^2 = \frac12 mv_{top}^2 + mg(2R)$$
$$v_{bot,\min} = \sqrt{v_{top,\min}^2 + 4gR} = \sqrt{gR+4gR} = \sqrt{5gR}$$
Tension at bottom with minimum speed:
$$T_{bot} = m\left(g + \frac{v_{bot}^2}{R}\right) = m\left(g + 5g\right) = 6mg$$
Answer
$$\boxed{T_{bot} = m\left(g + \frac{v_{bot}^2}{R}\right) = m\left(g + 5g\right) = 6mg}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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