Two metal disks, one with radius R 1 = 2.44cm and mass M 1 = 0.830kg and the oth

Question

Two metal disks, one with radius R1 = 2.44cm and mass M1 = 0.830kg and the other with radius R2 = 4.99cm and mass M2 = 1.63kg , are welded together and mounted on a frictionless axis through their common center.

Part A

What is the total moment of inertia of the two disks?

Part B

A light string is wrapped around the edge of the smaller disk, and a 1.50-kg block, suspended from the free end of the string. If the block is released from rest at a distance of 2.03m above the floor, what is its speed just before it strikes the floor?

Part C

Repeat the calculation of part B, this time with the string wrapped around the edge of the larger disk.

Explanation / Answer

here,

(A)

we know

moment of inertia for a disk = 0.5 * M * R^2

moment of inertia = moment of inertia of disk 1 + moment of inertia of disk 2

total moment of inertia = 0.5*M1 * R1^2 +0.5 * M2* R2^2

I(total) = 0.5(0.83*0.024^2 + 1.63 * 0.049^2)

I = 0.0021 kg m^2

total moment of inertia is 0.0021 kg m^2

(B)

using conservation of energy

we get

0.5 * I * w^2 + 0.5 * m *v^2 = mgh.......(1)

and we know

w v/r ... (2)

then

0.5 * 0.0021 * (v/0.024)^2 + 0.5 * 1.5 * v^2 = 1.5 * 9.8 * 2.03

v = 3.40m/s

speed of the block just above the ground is 3.40 m/s

(C)

when string is wrraped around larger disk

using coservation of enrgy

0.5 * 0.0021 * (v/0.049)^2 + 0.5 * 1.5 * v^2 = 1.5 * 9.8 * 2.03

v = 5.01 m/s

speed of block above the ground is 5.01 m/s

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