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The bob of a pendulum has its rest point...

The bob of a pendulum has its rest point 1 m below the support. The bob is pulled aside until the string makes an angle of `60^@` with the vertical line. On release with what speed does the bob swing pass its rest point?

A

`3.2m//s`

B

`4m//s`

C

`2m//s`

D

`5m//s`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will use the principle of conservation of mechanical energy. The potential energy at the highest point (when the bob is pulled aside) will convert into kinetic energy at the lowest point (the rest point). ### Step-by-Step Solution: 1. **Identify the Points**: - Let point A be the position of the bob when it is pulled to an angle of 60 degrees. - Let point B be the position of the bob at the rest point (the lowest point). 2. **Determine the Height Change**: - The length of the pendulum (L) is 1 m. - When the pendulum is pulled to 60 degrees, we can find the height (h) it has been raised using the cosine of the angle: \[ h = L - L \cos(60^\circ) = 1 - 1 \cdot \cos(60^\circ) = 1 - 1 \cdot \frac{1}{2} = 1 - 0.5 = 0.5 \text{ m} \] 3. **Calculate the Potential Energy at Point A**: - The potential energy (PE) at point A is given by: \[ PE_A = mgh = mg \cdot 0.5 \] - Here, \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)). 4. **Calculate the Kinetic Energy at Point B**: - At point B, all the potential energy will convert into kinetic energy (KE): \[ KE_B = \frac{1}{2} mv^2 \] 5. **Apply Conservation of Energy**: - According to the conservation of energy: \[ PE_A = KE_B \] - Substituting the expressions for potential and kinetic energy: \[ mg \cdot 0.5 = \frac{1}{2} mv^2 \] 6. **Cancel the Mass (m)**: - Since mass appears on both sides, we can cancel it: \[ g \cdot 0.5 = \frac{1}{2} v^2 \] 7. **Solve for v**: - Rearranging gives: \[ v^2 = g \] - Substituting \( g = 10 \, \text{m/s}^2 \): \[ v^2 = 10 \] - Taking the square root: \[ v = \sqrt{10} \approx 3.16 \, \text{m/s} \] 8. **Final Answer**: - Rounding off, we find that the speed of the bob as it passes the rest point is approximately: \[ v \approx 3.2 \, \text{m/s} \]
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Knowledge Check

  • The mass of a simple pendulum bob is 100 gm. The length of the pendulum is 1 m. The bob is drawn aside from the equilibrium position so that the string makes an angle of 60^(@) with the vertical and let go. The kinetic energy of the bob while crossing its equilibrium position will be

    A
    `0.49 J`
    B
    `0.94 J`
    C
    `1 J`
    D
    `1.2 J`

  • A bob is suspended from an ideal string makes an angle 60^(@) to the verticle and whirled along a horizontal circle.Then its perod of revolution is

    A
    `pi sqrt((l)/(g))`
    B
    `pisqrt((l)/(2g))`
    C
    `pi sqrt((2l)/(g))`
    D
    `2pisqrt((l)/(g))`

  • A pendulum bob on a 2 m string is displaced 60^(@) from the vertical and then released. What is the speed of the bob as it passes through the lowest point in its path

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    `sqrt(2)m//s`
    B
    `sqrt(9.8) m//s`
    C
    `4.43 m//s`
    D
    `1//sqrt(2) m//s`

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