Query 32

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course PHY 231

11/24 2:05 pm I've finished collecting results for the gravitation exp finally but won't be able to start analysis until next Tuesday b/c of other work such as this test and Dynamics test.

032. `query 32*********************************************

Question: `qUniv. 10.64 (10.56 10th edition). disks 2.5 cm and .8 kg, 5.0 cm and 1.6 kg, welded, common central axis. String around smaller, 1.5 kg block suspended. Accel of block? Then same bu wrapped around larger.

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Your solution:

This threw me at first bc I didn't think about how the block was adding to I.

(a)

I= 1/2(0.8)(0.025)^2 + 1/2(1.6)(0.05)^2+1.5(.025)^2 = 0.0032

tau= F*R1 = 1.5*9.81*0.02= 0.368

alpha = tau/I = 0.368/0.032= 115 rad/s^2

convert to linear accel:

a= alpha*R1= 115*0.025= 2.88 m/s^2

(b)

I= 1/2(0.8)(0.025)^2 + 1/2(1.6)(0.05)^2+1.5(0.05)^2 = 0.006

tau= F*R2= 1.5*9.81*0.05= 0.736

alpha= tau/I= 0.736/0.006 = 122.7 rad/s^2

a= alpha*R2= 122.7*0.05 = 6.14 m/s^2

Clearly b is larger and it is expected to be so since the larger R creates more torque for nearly the same I so alpha increases.

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Given Solution:

`a** The moment of inertia of each disk is .5 M R^2; the block lies at perpendicular distance from the axis which is equal to the radius of the disk to which it is attached. So the moment of inertia of the system, with block suspended from the smaller disk, is

I = .5 (.8 kg) * ( .025 m)^2 + .5 * 1.6 kg * (.05 m)^2 + 1.5 kg * (.025 m)^2= .0032 kg m^2 approx.

The 1.5 kg block suspended from the first disk results in torque

tau = F * x = .025 m * 1.5 kg * 9.8 m/s^2 = .37 m N approx.

The resulting angular acceleration is

alpha = tau / I = .37 m N / (.0032 kg m^2) = 115 rad/s^2 approx.

The acceleration of the block is the same as the acceleration of a point on the rim of the wheel, which is

a = alpha * r = 115 rad/s^2 * .025 m = 2.9 m/s^2 approx.

The moment of inertia of the system, with block suspended from the larger disk, is

I = .5 (.8 kg) * ( .025 m)^2 + .5 * 1.6 kg * (.05 m)^2 + (1.5 kg * .05 m)^2= .006 kg m^2 approx.

The 1.5 kg block suspended from the first disk results in torque

tau = F * x = .05 m * 1.5 kg * 9.8 m/s^2 = .74 m N approx.

The resulting angular acceleration is

alpha = tau / I = .74 m N / (.006 kg m^2) = 120 rad/s^2 approx.

The acceleration of the block is the same as the acceleration of a point on the rim of the wheel, which is

a = alpha * r = 120 rad/s^2 * .05 m = 6 m/s^2 approx. **

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Self-critique (if necessary):OK

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&#Good responses. Let me know if you have questions. &#

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