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nikklg [1K]
3 years ago
11

A 2.30-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. a solid sph

ere of mass 0.25 kg is thrown horizontally with a speed v1 = 12.5 m/s to hit the rod at the point a one-fifth of the way up from the bottom of the rod. the sphere bounces back horizontally with a speed v2 = 9.50 m/s, while the rod swings to the right through an angle θ before swinging back toward its original position. what is the angular speed of the rod immediately after the collision?
Physics
1 answer:
Marina86 [1]3 years ago
8 0

Solution:


initial sphere mvr = final sphere mvr + Iω 
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m² 
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω 
where: ω = 2.87 rad/s 

So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)² 
E = 12.64 J becomes PE = mgh, so 
12.64 J = 2.3 kg * 9.8m/s² * h 
h = 0.29 m 

h = L(1 - cosΘ) → where here L is the distance to the CM 
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ 
Θ = arccos((1-0.29)/1) = 44.77 º 

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Therefore, if we want to calculate luminosity the Joule as a unit will be used.

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If we input the units we will have:

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Where:

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3 years ago
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8 0
3 years ago
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A hot-air balloon and its basket are accelerating upward at 0.265 m/s2, propelled by a net upward force of 688 N. A rope of negl
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Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>

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The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.

Explanation:

Let us call the mass of the balloon m_1 and the mass of the basket m_2, then according to newton's second law:

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\boxed{m_2 = 301.13kg}

Putting this into equation (1) and substituting the numerical values of F and a, we get:

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Thus, the mass of the balloon and the basket is  2295 kg and 301 kg respectively.

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