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azamat
3 years ago
5

A 300 g ball and a 600 g ball are connected by a 40-cm-lon massless, rigid rod. The structure rotates about its center of me at

100 rpm. What is its rotational kinetic energy?

Physics
1 answer:
Readme [11.4K]3 years ago
7 0

Answer:

 KE = 1.75 J

Explanation:

given,

mass of ball, m₁ = 300 g = 0.3 Kg

mass of ball 2, m₂ = 600 g = 0.6 Kg

length of the rod = 40 cm = 0.4 m

Angular speed = 100 rpm= 100\times \dfrac{2\pi}{60}

                         =10.47\ rad/s

now, finding the position of center of mass of the system

    r₁ + r₂ = 0.4 m.....(1)

 equating momentum about center of mass

  m₁r₁ = m₂ r₂

   0.3 x r₁ = 0.6 r₂

   r₁ = 2 r₂

Putting value in equation 1

2 r₂ + r₂ = 0.4

 r₂ = 0.4/3

 r₁ = 0.8/3

now, calculation of rotational energy

KE = \dfrac{1}{2}I_1\omega^2+\dfrac{1}{2}I_2\omega^2

KE = \dfrac{1}{2}\omega^2 (I_1 +I_2)

KE = \dfrac{1}{2}\omega^2 (m_1r_1^2 +m_2r^2_2)

KE = \dfrac{1}{2}\times 10.47^2(0.3\times (0.8/3)^2 +0.6\times (0.4/3)^2)

 KE = 1.75 J

the rotational kinetic energy is equal to 1.75 J

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In Fig. 4-41, a ball is thrown up onto a roof, landing 4.00 s later at height h ???? 20.0 m above the release level. The ball’s
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"Fig is attacted with answer"

Answer:

a) d = 33.72 m

b) v_{i} = 26 m/s

c) β = 71.08°

Explanation:

a)

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

time = t = 4.00 s

Height = h = 20 m

Angle = θ = 60°

Horizontal distance = d = ?

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h = v_{y_{f}}t + \frac{1}{2}gt^{2}

-20 = v_{y_{f}} (4) + 0.5(-9.8)(4)²

v_{y_{f}} (4) = 58.4

v_{y_{f}}  = 14.6 m/s

This is vertical component of velocity when the ball is on the roof. To calculate the Final velocity and horizontal component, we use

v_{f} = v_{y_{f}} / sinθ

v_{f} = 14.6 / sin 60

v_{f} = 16.86 m/s

v_{x_{f}} = v_{f}cosθ

v_{x_{f}} = 16.86 cos 60

v_{x_{f}} = 8.43 m/s

To calculate the horizontal distance

d = v_{y_{f}} t

d = (8.43)(4)

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b)

We know the values of Landing angle, height of roof, time of flight. In part a, We calculate the landing velocity of the ball and also its horizontal and vertical component. As the ball followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

So,

v_{x_{f}} = v_{x_{i}}

but the vertical component of velocity vary with and there is an acceleration along vertical direction which is equal to gravitation acceleration g.

So,

g = (v_{y_{f}} - v_{y_{i}} ) / t

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v_{y_{i}} = 24.6 m/s

v_{i} = \sqrt{v_{x_{i}}^{2}+v_{y_{i}}^{2} }

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v_{i} = 26 m/s

c)

cos β = v_{x_{i}} / v_{i}

β = cos⁻¹ (8.43 / 26)

β = 71.08°

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