Question

An object is released from rest from the inner edge of a hemispherical bowl, and it falls under gravity. The coefficient of kinetic friction between the object and the bowl is $\mu$. If the object covers an angular displacement $\theta$ with respect to the center of the hemisphere when it stops for the first time, which of the following expressions is correct?

• A. $\mu = cot(\theta/2)$
• B. $\mu = cot(\theta)$
• C. $\mu = tan(\theta)$
• D. $\mu = tan(\theta/2)$

Answer 1

Answer: (A)

$\mu = cot(\theta/2)$

Answer 2

Assumptions:

• The object starts from the rim of the hemisphere.
• We approximate the normal force $N \approx mg \cos \alpha$, neglecting the $mv^2/R$ term.

Derivation:

1. Work done by gravity ($W_g$):

   • Initial height: $h_i = 0$
   • Final height: $h_f = -R \sin \theta$
   • $W_g = mg(h_i - h_f) = mgR \sin \theta$

2. Work done by friction ($W_f$):

   • $W_f = \int_{\pi/2}^{\pi/2-\theta} -\mu N R d\alpha$
   • Using $N \approx mg \cos \alpha$:
   • $W_f \approx -\mu mgR \int_{\pi/2}^{\pi/2-\theta} \cos \alpha d\alpha = -\mu mgR (\cos \theta - 1)$
   • $W_f = \mu mgR (1 - \cos \theta)$

3. Work-energy theorem:

   • $W_g + W_f = 0$
   • $mgR \sin \theta + \mu mgR (1 - \cos \theta) = 0$
   • $\mu (1 - \cos \theta) = \sin \theta$
   • $\mu = \frac{\sin \theta}{1 - \cos \theta}$

4. Simplifying using trigonometric identities:

   • $\mu = \frac{2 \sin(\theta/2) \cos(\theta/2)}{2 \sin^2(\theta/2)} = \cot(\theta/2)$

Therefore, $\mu = \cot(\theta/2)$.