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3 years ago

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A woman rides a carnival Ferris wheel at radius 20 m, completing 6.0 turns about its horizontal axis every minute. What are (a)

the period of the motion, and the magnitude of her centripetal acceleration at (b) the highest point and (c) the lowest point?

1 answer:

natta225 [31]3 years ago

7 0

Answer:

Explanation:

Given

radius of ferris wheel is 20 m

It completes 6 turns in 1 min

i.e. 1 turn in 10 sec

Therefore its angular velocity is $\omega =\frac{2\pi }{10} rad/s$

(a)Period of motion is 10 s

Magnitude of centripetal acceleration is $\omega ^2r$

$=\left ( \frac{\pi }{5}\right )^2\times 20=7.89 m/s^2$

(b)Highest point will be 40 m

(c) lowest point 0 m i.e. at ground

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The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g

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Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:

$k=\sqrt{\frac{I}{M} }$

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration $k=\sqrt{\frac{I}{M} }$

So, moment of rotational inertia (I) of a cylinder about it axis = $\frac{MR^2}{2}$

$k=\sqrt{\frac{\frac{MR^2}{2}}{M} }$

$k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }$

$k=\sqrt{{\frac{R^2}{2}}$

$k={\frac{R}{\sqrt{2}}$

$k={\frac{1.20m}{\sqrt{2}}$

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = $\frac{2}{3}MR^2$

$k = \sqrt{\frac{\frac{2}{3}MR^2}{M} }$

$k = \sqrt{\frac{2}{3} R^2}$

$k = \sqrt{\frac{2}{3} }*R$

$k = \sqrt{\frac{2}{3}} *1.20$

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of radius

(I) = $\frac{2}{5}MR^2$

$k = \sqrt{\frac{\frac{2}{5}MR^2}{M} }$

$k = \sqrt{\frac{2}{5} R^2}$

$k = \sqrt{\frac{2}{5} }*R$

$k = \sqrt{\frac{2}{5}} *1.20$

k = 0.7560

k ≅ 0.76 m

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