Question

25. In a Circus, a motor-cyclist having mass of 50 kg moves in a spherical cage of radius 3 m. Calculate the least velocity with which he must pass the highest point without losing contact. Also calculate his angular speed at the highest point.​

Answer 1

The angular speed at the highest point is ω = 1.807 rad / s

Explanation:

Force acting on motorcyclist = mg + N

When motorcyclist just lose contact. N = 0

mv^2 / r = mg

"m" gets cancelled on both sides.

v^2 = rg

v = √ rg

v = √  3 x 9.8

v = 5.421 m/s

ω = v / r

ω = 5.421 / 3 = 1.807 rad / s

Thus the angular speed at the highest point is ω = 1.807 rad / s

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Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

Given :

• Mass (m) = 50 kg
• Radius of Road (r) = 3 m

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To Find :

• Angular Speed at highest Point

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Solution :

As we know that :

$\large{\boxed{\sf{F \: = \: \dfrac{mv^2}{r}}}}$

And also, F = mg

So,

$\implies {\sf{\cancel{m}g \: = \: \dfrac{\cancel{m}v^2}{r}}} \\ \\ \implies {\sf{g \: = \: \dfrac{v^2}{r}}} \\ \\ \implies {\sf{v^2 \: = \: rg}} \\ \\ \implies {\sf{v^2 \: = \: 3 \: \times \: 10}} \\ \\ \implies {\sf{v^2 \: = \: 30}} \\ \\ \implies {\sf{v \: = \: \sqrt{30}}} \\ \\ {\sf{v \: = \: 5.47}} \\ \\ {\underline{\sf{\therefore \: Velocity \: is \: 5.47 \: ms^{-1}}}}$

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We have formula for Angular Velocity :

$\large {\boxed{\sf{\omega \: = \: \dfrac{v}{r}}}} \\ \\ \implies {\sf{\omega \: = \: \dfrac{5.47}{3}}} \\ \\ \implies {\sf{\omega \: = \: 1.82}} \\ \\ {\underline{\sf{\therefore \: Angular \: velocity \: at \: heighest \: point \: is \: 1.82 \: rads^{-1}}}}$