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bija089 [108]
3 years ago
5

A 103 kg horizontal platform is a uniform disk of radius 1.71 m and can rotate about the vertical axis through its center. A 68.

9 kg person stands on the platform at a distance of 1.09 m from the center, and a 27.7 kg dog sits on the platform near the person 1.45 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.
Physics
1 answer:
Andreyy893 years ago
5 0

Answer:

I_{total}=220.64 kg*m^{2}

Explanation:

The moment of inertia of the system is equal to the each population and the platform inertia so

Inertia disk

I_{disk}=\frac{1}{2}*m_{disk}*(r_{p})^{2}

Inertia person

I_{p}=\frac{1}{2}*m_{p}*(r_{p})^{2}

Inertia dog

I_{d}=\frac{1}{2}*m_{d}*(r_{d})^{2}

The Inertia of the system is the sum of each mass taking into account that all exert the force of inertia:

I_{total}=I_{disk}+I_{p}+I_{d}

I_{total}=\frac{1}{2}*103kg*(1.71)^{2}+\frac{1}{2}*68.9kg*(1.09)^{2}+\frac{1}{2}*27.7kg*(1.45)^{2}

I_{total}=220.64 kg*m^{2}

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A playground is on the flat roof of a city school, 6.2m above the street below. The vertical wall of the building is h=7.30m hig
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Answer:

a. 18.13m/s

b. 0.84m

c. 2.4m

Explanation:

a. to find the speed at which the ball was lunched, we use the horizontal component.Since the point distance from the base of the ball is 24m and it takes 2.20 secs to reach the wall,we can say that

t=distance /speed

speed v_{x}=vcos\alpha=24/2.2=10.9m/s\\V=10.9/cos53^{0}\\V=18.13m/s

Hence the speed at which ball was lunched is 18.13m/s

b. from the equation

y=v_{y}t-\frac{1}{2}gt^{2}\\ v_{y}=18.13sin53=14.48m/s\\\\y=(14.48*2.2)-\frac{1}{2}9.8*2.2^{2}\\y=8.14m\\

the vertical distance at which the ball clears the wall is

y=8.14-7.3=0.84m

c. the time it takes the ball to reach the 6.2m vertically

6.2=14.47t-4.8t^{2}\\\\14.47t-4.8t^{2}-6.2=0\\\using\\t=-b±\frac{\sqrt{b^{2}-4ac} }{2a}\\ where \\a=-4.8, b=14.47t, c=6.2\\hence t=2.4secs

the horizontal distance covered at this speed is

y+24=(18.13cos53)2.4\\y=26.186-24\\y=2.12m

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