Uncategorized: Practice Problem & Solution
A particle moves along a circle of radius \( left( frac{20}{pi} right) text{ m} \) with constant tangential acceleration. If the velocity of the particle is \( 80 text{ m/s} \) at the end of the second revolution after motion has begun, the tangential acceleration is: (2003)
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Uncategorized. Understanding the underlying formula is key to arriving at the correct answer below:
Given: \( r = frac{20}{pi} text{ m} \), \( v = 80 text{ m/s} \). The angular displacement is \( theta = 2 times 2pi = 4pi text{ rad} \). The distance covered is \( s = rtheta = frac{20}{pi} times 4pi = 80 text{ m} \). Using the kinematic equation \( v^2 = u^2 + 2as \), we get \( (80)^2 = 0 + 2 a_t (80) \), which yields \( a_t = 40 text{ m/s}^2 \).
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