Question

A Car is moving in a circular horizontal
track of radius 10m with a constant
speed of 10 m/s. A plumb bob is
suspended from the roof of the car
by a light rod. The angle made by the
rod with the vertical is (Take g = 10 m/s²)​

Answer 1

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A car is moving in a circular horizontal track of radius 10m with a constant speed of 10m/s. A pendulum bob is suspended from the roof of the car by a light, rigid support rod of length 1m. What would be the angle made by the rod with the vertical is ? (Take g = 10 m/s²).

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Given Radius(r) = 10 m.
• Given Speed(v) = 10 m/s.
• Acceleration due to gravity (g) = 10 m/s².

$\huge{\bold{\underline{Explanation:-}}}$

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From the figure we can see that "AB" is the rod , and the bob is suspended from Point "A".

Now, Let the bob has a mass "m", its weight will act as "mg",

Let the Tension in the rod be "T".

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Now, The system should be in equilibrium,

So, Resolving the Tension "T" in to components,

This will give two components of Tension I.e.

one Tcosθ and another one Tsinθ.

Now, Applying the concept of Equilibrium,

$\large{\tt \leadsto Tcos \theta = mg \: -----(1)}$

And other one as,

$\large{\tt \leadsto Tsin \theta = \dfrac{mv^2}{r} \: -----(2)}$

Now, Dividing equation (1)/equation (2).

$\large{\tt \leadsto \dfrac{T cos \theta}{T sin \theta} = \dfrac{mg}{\dfrac{mv^2}{r}}}$

$\large{\tt \leadsto \dfrac{\cancel{T} cos \theta}{\cancel{T} sin \theta} = \dfrac{mg}{\dfrac{mv^2}{r}}}$

It becomes,

$\large{\tt \leadsto \dfrac{cos \theta}{sin \theta} = \dfrac{mg \times r}{mv^2}}$

$\large{\tt \leadsto Tan \theta = \dfrac{\cancel{m}g \times r}{\cancel{m}v^2}}$

$\large{\tt \leadsto Tan \theta = \dfrac{g \times r}{v^2}}$

Substituting the values,

$\large{\tt \leadsto Tan \theta = \dfrac{10 \times 10}{(10)^2}}$

$\large{\tt \leadsto Tan \theta = \dfrac{100}{100}}$

$\large{\tt \leadsto Tan \theta = \dfrac{\cancel{100}}{\cancel{100}}}$

$\large{\tt \leadsto Tan \theta = 1}$

$\large{\tt \leadsto Tan \theta = Tan 45\degree}$

$\huge{\boxed{\boxed{\tt \theta = 45 \degree}}}$

So, the Angle made by the rod with vertical is 45°.

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#refer the attachment for figure.

(Figure is the F.B.D of the rod and bob)