Circular Motion #4

A particle moves in a circular path of radius 1 1 . Its speed v v , at time t t is given by v = 5 + 4 sin ( π 6 t ) v = 5 + 4 \sin ( \frac{ \pi }{ 6 } t ) . What is the magnitude of the tangential acceleration of the particle at t = 5 t = 5 ?

The given information is in SI units. State your answer in SI units too.


This problem is part of the set - Circular Motion Practice


The answer is 1.814.

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1 solution

Pranshu Gaba
Nov 4, 2015

The tangential acceleration of a particle moving in a circular path is d v d t \frac{ dv } {dt } , that is the derivative of speed of the particle with respect to time. In the problem,

a T = d v d t = 4 π 6 cos ( π 6 t ) a_{T} = \frac{ dv } { dt} = \frac{ 4 \pi } { 6 } \cos \left( \frac{\pi } { 6 } t\right)

When t = 5 t = 5 , a T = 4 π 6 cos ( 5 π 6 ) = 4 π 6 × 3 2 a_{T} = \dfrac{ 4 \pi } { 6 } \cos \left( \dfrac{5 \pi } { 6 } \right) = \dfrac{ 4 \pi } { 6 } \times \dfrac{ - \sqrt {3} } { 2} .

The magnitude of tangential acceleration is a T = 4 π 6 × 3 2 1.814 | a_{T} | = \dfrac{ 4 \pi } { 6 } \times \dfrac{\sqrt {3} } { 2} \approx \boxed{ 1.814 } _\square

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