Laws of Motion 4 Question 6
6. A car is moving in a circular horizontal track of radius $10 \mathrm{~m}$ with a constant speed of $10 \mathrm{~m} / \mathrm{s}$. A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is (Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )
$(1992,2 \mathrm{M})$
(a) zero
(b) $30^{\circ}$
(c) $45^{\circ}$
(d) $60^{\circ}$
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Answer:
Correct Answer: 6. (c)
Solution:
- FBD of bob is $T \sin \theta=\frac{m v^{2}}{R}$
and $T \cos \theta=m g$
$\therefore \tan \theta=\frac{v^{2}}{R g}=\frac{(10)^{2}}{(10)(10)}$
$$ \tan \theta=1 $$
$$ \text { or } \quad \theta=45^{\circ} $$
