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3 years ago

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A girl of mass 50.6 kg stands on the edge of a frictionless merry-go-round of mass 827 kg and radius 3.72 m that is not moving.

She throws a 1.13-kg rock in a horizontal direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is 7.82 m/s. Calculate (a) the angular speed of the merry-go-round and (b) the linear speed of the girl after the rock is thrown. Assume that the merry-go-round is a uniform disk.

1 answer:

GaryK [48]3 years ago

7 0

Answer:

(a). The value of angular speed of the merry-go-round $\omega$ = - 5.82 × $10^{-3}$ $\frac{rad}{s}$

(b). The linear speed of the girl after the rock is thrown V = -1.89 × $10^{-2}$ $\frac{m}{s}$

Explanation:

Given data

Mass of the girl $m_{g}$ = 50.6 kg

Mass of merry-go-round $m_{m}$ = 827 kg

Radius r = 3.72 m

The speed of the rock relative to the ground $V_{r}$ = 7.82 $\frac{m}{s}$

(a). The angular speed of the merry-go-round is given by

$\omega = - [\frac{m_{r}v_{r} }{r} ] \frac{2}{m_{m} + 2m_{g} }$

Put all the values in above formula

$\omega = \frac{(1.13)(7.82)}{3.27} \frac{2}{827 + (2)50.6}$

$\omega$ = - 5.82 × $10^{-3}$ $\frac{rad}{s}$

This is the value of angular speed of the merry-go-round.

(b). The liner speed of the girl is given by

⇒ V = r × $\omega$

⇒ V = - 3.72 × 5.82 × $10^{-3}$

⇒ V = -1.89 × $10^{-2}$ $\frac{m}{s}$

This is the linear speed of the girl after the rock is thrown.

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Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

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<u>Given:</u>

Time of flight = t = 20 s
Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

<u>Assume:</u>

Initial velocity of the projectile = u
R = Range of the projectile during the time of flight
H = maximum height of the projectile
D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

Initial horizontal velocity of the projectile = $u\cos \theta$
Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

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Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

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Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

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