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Circular Motion

Section: Physics

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Question : 61 of 76

Marks: +1, -0

A block of

200\ \text{g}

mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius

20\ \text{cm}

. If the block takes 40 s to complete one round, the normal force by the side walls of the groove is

[16 Mar 2021 Shift 1]

$0.0314\ \text{N}$
$9.859 \times 10^{-2}\ \text{N}$
$6.28 \times 10^{-3}\ \text{N}$
$9.859 \times 10^{-4}\ \text{N}$

Solution:

The normal force by the side walls of the groove will be equal to the centripetal force acting on it.

i.e.

N = \frac{m v^{2}}{r}

.....(i)

where,

r = 20\ \text{cm} = 0.2\ \text{m}

m = 200\ \text{g} = 200 \times 10^{-3}\ \text{kg}

and

v = r \omega = \frac{2 \pi r}{T} = \frac{2 \pi \times 0.2}{40}\ \text{m/s}

Substituting the given values in Eq. (i), we get

N = \frac{(200 \times 10^{-3}) \times \left(\frac{2 \pi \times 0.2}{40}\right)^{2}}{0.2}

\approx 9.859 \times 10^{-4}\ \text{N}

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