# As the string passes through the vertical, the narrow beam of light is interrupted by the cylinder for 22 ms . The cylinder has a diameter of 2.5 cm. How to calulate the actual speed of the cylinder?

Jul 6, 2018

Distance moved by the cylinder as it interrupts the beam of light $= \text{Diameter of cylinder} = 2.5 \setminus c m$
Time taken to cover this distance $= 22 \setminus m s$
Average speed of the cylinder ${v}_{a} = \text{Distance"/"Time}$

$\implies {v}_{a} = \frac{\frac{2.5}{100}}{22 \times {10}^{-} 3}$
$\implies {v}_{a} = 1.14 \setminus m {s}^{-} 1$, rounded to two decimal places.

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Assuming that full question is Question 1 part (c)i

(b) $\frac{1}{2} m {v}^{2} = m g h$
$\implies v = \sqrt{2 g h}$
$\implies v = \sqrt{2 \times 10 \times \left(2 - 1.932\right)} = 1.17 \setminus m {s}^{-} 1$