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FinnZ [79.3K]
4 years ago
15

A potter's wheel has the shape of a solid uniform disk of mass 7 kg and radius 0.65 m. It spins about an axis perpendicular to t

he disk at its center. A small 2.1 kg lump of very dense clay is dropped onto the wheel at a distance 0.41 m from the axis.
What is the moment of inertia of the system about the axis of spin?
Physics
1 answer:
ozzi4 years ago
5 0

Answer:

1.832 kgm^2

Explanation:

mass of potter's wheel, M = 7 kg

radius of wheel, R = 0.65 m

mass of clay, m = 2.1 kg

distance of clay from centre, r = 0.41 m

Moment of inertia = Moment of inertia of disc + moment f inertia of the clay

I = 1/2 MR^2 + mr^2

I = 0.5 x 7 x 0.65 x 0.65 + 2.1 x 0.41 x 0.41

I = 1.47875 + 0.353

I = 1.832 kgm^2

Thus, the moment of inertia is 1.832 kgm^2.

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Speed of Dayshawn travelling towards his home is 12 m/s

Speed of her mom towards his school is 5 m/s

They both starts at same time so whenever they will meet on their path the sum of the distance covered by Dayshawn and distance covered by his mom must be equal to the total distance of school and home

Now let say they both meet after "t" time when they starts motion

so we can write the total distance between school and home as

d = v_1*t + v_2*t

here d = 6492 m

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now by solving the above equation

6492 = 12t + 5 t

t = \frac{6492}{17}

t = 381.9 s

so they will meet after 381.9 s from start which will be 3.36 minutes from there start

Also at this time the distance covered by her mom will be

d_2 = v_2*t

d_2 = 5* 381.9 = 1909.4 m

so they will meet at distance 1909.4 m from their home

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