Question

A particle moves in a circle of radius

2.0 cm

at a speed given by,

v = 4 t

, where

v

is in

cm s^{- 1}

and

t

is in seconds.
(a) Find the tangential acceleration at

t = 1 s

.
(b) Find total acceleration at

t = 1 s

.

Accepted Answer

Recall the formula of tangential and centripetal accelerations,

$a_{T} = \frac{d v}{d t}$ and $a_{c} = \frac{v^{2}}{r}$ .

Here, $v and r$ are velocity of the particle and the radius of the circle in which the particle is moving.

(a) Tangential acceleration, $t = 1 s$
$\Rightarrow a_{T} = \frac{d v}{d t} = \frac{d (4 t)}{d t} = 4 cm s^{- 2}$ .

(b) The value of centripetal acceleration,

$a_{c} = \frac{v^{2}}{r}$ .

Here, $v = 4 t$ and at $t = 1 s$ ,

$\Rightarrow a_{c} = \frac{{(4 t)}^{2}}{2} = 8 cm s^{- 2}$ .

As tangential and centripetal accelerations are normal to each other, so the magnitude of acceleration, say total acceleration,

$a_{total} = \sqrt{{a_{T}}^{2} + a_{C}^{2}} = \sqrt{(4)^{2} + (8)^{2}} = 4 \sqrt{5} cm s^{- 2}$ .

B M Sharma Solutions for Chapter: Circular Motion, Exercise 15: ILLUSTRATION 9.15

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Circular Motion, Exercise 15: ILLUSTRATION 9.15

Questions from B M Sharma Solutions for Chapter: Circular Motion, Exercise 15: ILLUSTRATION 9.15 with Hints & Solutions

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