Question

A particle moving along a circular path of radius 5 m with a uniform spees is 55 m/s .th aberage acceleration when the particle completes half revolution iS

Answer 1

Answer:

Given:

Radius = 5 metres

Uniform speed = 55 m/s

To find:

Average acceleration to complete half revolution.

Concept:

In case of uniform circular motion, the only acceleration experienced by the particle is the centripetal acceleration.

Calculation:

Centripetal force

$= \dfrac{m {v}^{2} }{r}$

Hence , centripetal acceleration

$= \frac{ (\frac{m {v}^{2} }{r} )}{m}$

$= \frac{ {v}^{2} }{r}$

$= \frac{ {(55)}^{2} }{5}$

$= 605 \: m \: {s}^{ - 2}$

So final answer is

$\boxed{ \red{average \: acc. = 605 \: m {s}^{ - 2}}}$

Answer 2

Answer:

$\large\boxed{ \sf{605 \: m {s}^{ - 2} }}$

Explanation:

Given that, a particle moves along a Circular path.

Radius of path, r = 5 m

Uniform speed, v = 55 m/s

We have to find the average acceleration.

• We know that, during uniform motion, the acceleration experienced by an object is only centripetal acceleration.

Now, we know that,

centripetal force = $\sf{\dfrac{m{v}^{2}}{r}}$

Therefore, we have,

Centripetal acceleration, $\sf{a =\dfrac{ \frac{m {v}^{2} }{r} }{m} =\dfrac{{v}^{2}}{r}}$

Therefore, we have , average acceleration,

$= > a = \frac{ {(55)}^{2} }{5} \\ \\ = > a = \frac{55 \times \cancel{ 55}}{ \cancel{5} } \\ \\ = > a = 55 \times 11 \\ \\ = > \sf{a = 605 \: m {s}^{ - 2} }$