A particle of mass $10 ext{ g}$ moves along a circle of radius $6.4 ext{ cm}$ with a constant tangential a

Question

A particle of mass $10 	ext{ g}$ moves along a circle of radius $6.4 	ext{ cm}$ with a constant tangential acceleration. What is the magnitude of this acceleration, if the kinetic energy of the particle becomes equal to $8 	imes 10^{-4} 	ext{ J}$ by the end of the second revolution after the beginning of the motion?

Accepted Answer

1. **Concept:** According to the **Work-Energy Theorem**, the work done by the net force on a particle is equal to the change in its kinetic energy ($W = \Delta K$) . In non-uniform circular motion, the tangential force $F_t$ does work over the distance covered.
2. **Formulas:**
   - Tangential Force: $F_t = m a_t$
   - Work Done: $W = F_t \cdot s = m a_t s$
   - Distance covered in $n$ revolutions: $s = n 	imes 2\pi R$
3. **Given Values:**
   - Mass ($m$) = $10 	ext{ g} = 10^{-2} 	ext{ kg}$
   - Radius ($R$) = $6.4 	ext{ cm} = 6.4 	imes 10^{-2} 	ext{ m}$
   - Kinetic Energy ($K$) = $8 	imes 10^{-4} 	ext{ J}$ (Initial $K_i = 0$)
   - Revolutions ($n$) = 2
4. **Calculation:**
   - Distance ($s$) = $2 	imes 2\pi 	imes (6.4 	imes 10^{-2}) = 4\pi (6.4 	imes 10^{-2}) 	ext{ m}$
   - Applying Work-Energy Theorem:
     $$ m a_t s = K $$
     $$ (10^{-2}) a_t [4\pi (6.4 	imes 10^{-2})] = 8 	imes 10^{-4} $$
     $$ a_t = \frac{8 	imes 10^{-4}}{10^{-2} 	imes 4\pi 	imes 6.4 	imes 10^{-2}} $$
     $$ a_t = \frac{8 	imes 10^{-4}}{4\pi 	imes 6.4 	imes 10^{-4}} = \frac{8}{25.6 \pi} $$
     $$ a_t = \frac{1}{3.2 \pi} \approx \frac{1}{3.2 	imes 3.1416} \approx \frac{1}{10.05} $$
     $$ a_t \approx 0.1 	ext{ m/s}^2 $$

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