Question

PartA A light string is wrapped around the outer rim of a solid uniform cylinder of diameter 75.0 cm that can rotate without friction about an axle through its center. A 3.00 kg stone is tied to the free end of the string, as shown in the figure (Figure 1) When the system is released from rest, you determine that the stone reaches a speed of 3.50 m/s after having fallien 2.50m What is the mass of the cylinder? kg Submit Request Answer Provide Feedback Figure 1 of 1 3.00 kg
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Answer #1

a = v2 - v02 / 2 d = (3.52 - 0) / (2 * 2.5)

a = 2.45 m/s2

T = m (g - a)

= 3 (9.8 - 2.45) = 22.05 N

torque = I * alpha

T R = (1/2 M R2) * (a / R)

mass of cyclinder = 2 T / a

= 2 * 22.05 / 2.45

mass of cyclinder = 18 kg

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