A 250-g weight is tied to a piece of thread wrapped around a spool, which is sus

Question

A 250-g weight is tied to a piece of thread wrapped around a spool, which is suspended in such a way that it can rotate freely. When the weight is released, it accelerates toward the floor as the thread unwinds. Assume that the spool can be treated as a uniform solid cylinder of radius R = 4 cm and mass M_s = 300 g. Find the tension in the thread and the magnitude of the acceleration of the weight as it descends. Assume the thread has negligible mass and does not slip or stretch as it unwinds. acceleration = tension =

Explanation / Answer

Is = k*mR² where k = 0.5
M = 0.250 kg
m = 0.300 kg

a = g[1 - 1/(M/km + 1)] = 6.125 m/s²

T = M*(g - a) = 0.91875 N

Or

T – Weight = mass * acceleration
Weight = 0.25 * 9.8 = 2.45 N
2.45 – T = 0.25 * a
T = 2.45 – 0.25 * a

For the spool, I drew circle with string wrapped around it. The tension is producing a torque that causes the spool to accelerate.

Torque = T * r = T * 0.04

AND

Torque = I *
I = ½ * m * r^2 = ½ * 0.3 * 0.04^2 = 2.4 * 10^-4
Since we need to determine the acceleration, let’s convert to a by dividing by r.
= a ÷ 0.04

Torque =2.4 * 10^-4 * (a ÷ 0.04) = a * 6 * 10^-3
Set the two torque equations equal to each other and solve for T.

T * 0.04 = a * 6 * 10^-3
T = a * 0.15
Set the two equation for T equal to each other and solve for a.

2.45 – 0.25 * a = a * 0.15
2.45 = a * 0.4
a = 6.125 m/s^2

To see if this is the correct acceleration, let’s substitute this number into the following equations and solve for T.

T = 2.45 – 0.25 * a = 2.45 – 0.25 * 6.125 = 0.91875 N
T = a * 0.15 = 6.125 * 0.15 = 0.91875 N
This proves that the tension and acceleration are correct.

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