To make a pendulum, a 210 g ball is attached to one end of a string that has a length of 1.2 m and negligible mass. (The other end of the st

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To make a pendulum, a 210 g ball is attached to one end of a string that has a length of 1.2 m and negligible mass. (The other end of the string is fixed.) The ball is pulled to one side until the string makes an angle of 24° with the vertical; then (with the string taut) the ball is released from rest. Find (a) the speed of the ball when the string makes an angle of 12° with the vertical and (b) the maximum speed of the ball. (c) What is the angle between the string and the vertical when the speed of the ball is one-third its maximum value?

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RobertKer 5 years 2021-07-22T08:15:19+00:00 1 Answers 25 views 0

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  1. trungdung22
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    2021-07-22T08:16:56+00:00
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    Answer:

    a) v \approx 1.233\,\frac{m}{s}, b) v \approx 1.426\,\frac{m}{s}, c) \theta_{2} \approx 22.61^{\textdegree}

    Explanation:

    a) The speed of the ball is determined by applying the Principle of Energy Conservation:

    U_{g,1} + K_{1} = U_{g,2} + K_{2}

    The speed of the ball when the string makes an angle of 12° with the vertical is:

    K_{2} = U_{g,1} - U_{g,2} + K_{1}

    \frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot L \cdot [(1-\cos \theta_{1})-(1 -\cos \theta_{2})]

    v = \sqrt{2\cdot g \cdot L \cdot (\cos \theta_{2} - \cos \theta_{1})}

    v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.2\,m)\cdot (\cos 12^{\textdegree}-\cos 24^{\textdegree})}

    v \approx 1.233\,\frac{m}{s}

    b) The maximum speed of the ball is:

    v = \sqrt{2\cdot g \cdot L \cdot (\cos \theta_{2} - \cos \theta_{1})}

    v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.2\,m)\cdot (\cos 0^{\textdegree}-\cos 24^{\textdegree})}

    v \approx 1.426\,\frac{m}{s}

    c) The angle between the string and the vertical when the speed of the ball is one-third its maximum value is obtained by proving different values of \theta_{2}. The solution is approximately:

    \theta_{2} \approx 22.61^{\textdegree}

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