A pendulum bob swings along a circular path on a smooth inclined plane as shown in Fig- 8-280- where m - 3 kg- l - 0-75 m- - - 370- At the lowest point of the circle the tension in the string is T - 274 N- Take g - 10 ms-1-The speed of the bob at the highest point on the circle is-

The correct option is A √(46) ms^ 1 All formula will be same as in circular motion in vertical #160;plane. Only replace, acceleration due to gravity (g) ...

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Standard XII

Physics

Centripetal Acceleration

A pendulum bo...

Question

A pendulum bob swings along a circular path on a smooth inclined plane as shown in Fig. 8.280, where m = 3 kg, l = 0.75 m, $θ = 37^{0}$ . At the lowest point of the circle the tension in the string is T = 274 N. Take g = 10 $m s^{- 1}$ .
The speed of the bob at the highest point on the circle is:

\sqrt{46} m s^{- 1}

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\sqrt{26} m s^{- 1}

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\sqrt{52} m s^{- 1}

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\sqrt{35} m s^{- 1}

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Solution

The correct option is A $\sqrt{46} m s^{- 1}$
All formula will be same as in circular motion in vertical plane. Only replace, acceleration due to gravity (g) by $g = g sin θ$ Due to wedge.
$T_{B o t t o m} = T_{T o p} + 6 m g$ (For vertical plane)
$T_{B o t t o m} = T_{T o p} + 6 m g s i n 37$ (For our case)
$∴ T_{T o p} = 274 - 6 \times 3 \times 10 \times sin 37$
$T_{T o p} = 165.67 \neq 166 N$
$T_{T o p} = \frac{m V^{2} T o p}{r} - m g^{'} \Rightarrow V_{T o p} = \sqrt{46} m / s$

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A pendulum bob swings along a circular path on a smooth inclined plane as shown in Fig. 8.280, where m = 3 kg, l = 0.75 m, θ=370. At the lowest point of the circle the tension in the string is T = 274 N. Take g = 10 ms−1.The speed of the bob at the highest point on the circle is:

A pendulum bob swings along a circular path on a smooth inclined plane as shown in Fig. 8.280, where m = 3 kg, l = 0.75 m, $θ = 37^{0}$ . At the lowest point of the circle the tension in the string is T = 274 N. Take g = 10 $m s^{- 1}$ .
The speed of the bob at the highest point on the circle is: