Chapter 10: Q.9 (page 256)

A pendulum is made by tying a $500 g$ ball to a $75 - c m - l o n g$ string. The pendulum is pulled $30 °$ to one side, then released. What is the ball’s speed at the lowest point of its trajectory?

Short Answer

Expert verified

The final answer speed = $0.81 m / s .$

Step by step solution

Introduction

As the bob goes from point D to E to F to G, the force and acceleration are directed rightward, and the velocity decreases. The pendulum bob has a velocity of 0 m/s at G, which is the greatest leftward displacement.

Given

There is a pendulum whose mass is = $500 g$

Converting gram into kg = $0.5 k g$

Length of string is = $75 c m$

75 cm into meter = $75 c m$

Pendulum pulled upto $30 °$

Explanation

By diagram

$h = l - l \cos 30 °$

$h = l (1 - \cos 30 °)$

$h = 0.5 (1 - \cos 30 °)$

$h = 0.65$

Now, change in potential energy

$= m g h$

$= 0.5 * 9.8 * 0.65$

$= 0.318 J$

Let ball speed at lowest point is L

so $\frac{1}{2} m v^{2} = 0.318$

$= \frac{1}{2} * 0.5 * v^{2} = 0.318$

$v = \sqrt{\frac{0.318 * 2}{0.5}}$

$v = 0.81 m / s$

The final answer speed = $0.81 m / s .$

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A pendulum is made by tying a $500 g$ ball to a $75 - c m - l o n g$ string. The pendulum is pulled $30 °$ to one side, then released. What is the ball’s speed at the lowest point of its trajectory?

Short Answer

Step by step solution

Introduction

Given

Explanation

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A pendulum is made by tying a500gball to a 75-cm-longstring. The pendulum is pulled30° to one side, then released. What is the ball’s speed at the lowest point of its trajectory?

Short Answer

Step by step solution

Introduction

Given

Explanation

One App. One Place for Learning.

Most popular questions from this chapter

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A pendulum is made by tying a $500 g$ ball to a $75 - c m - l o n g$ string. The pendulum is pulled $30 °$ to one side, then released. What is the ball’s speed at the lowest point of its trajectory?