# If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

The two blocks in (Figure 1) are connected by a massless rope that passes over a pulley. The pulley is 12 cm in diameter and has a mass of 1.9 kg . As the pulley turns, friction at the axle exerts a torque of magnitude 0.60 N⋅m

Relevance

net torque on pulley

τ = [4.0kg*(9.8m/s² - a) - M*(9.8m/s² + a)]*0.06m - 0.60N·m

where M is the smaller mass.

and τ = I*α

[4.0kg*(9.8m/s² - a) - M*(9.8m/s² + a)]*0.06m - 0.60N·m =

½*1.9kg*(0.06m)² * α

α = ([4.0kg*(9.8m/s² - a) - M*(9.8m/s² + a)]*0.06m - 0.60N·m) / 0.00342kg·m²

so the linear acceleration

a = α*r = α * 0.06m

a = ([4.0kg*(9.8m/s² - a) - M*(9.8m/s² + a)]*0.06m - 0.60N·m) / 0.057kg·m

and then

t = √(2a / h)

where h is the height

FOR INSTANCE, if M = 2.0 kg and h = 1.0 m, then

a = ([4.0kg*(9.8m/s² - a) - 2.0kg*(9.8m/s² + a)]*0.06m - 0.60N·m) / 0.057kg·m

solves to

a = 1.4 m/s²

in which case

t = √(2 * 1.0m / 1.38m/s²) = 1.2 s

• Good!  I failed to get the radius correct. DOH!!!

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• Let's assume , in lack of the figure, that the two masses are worth 4 and 2 kg ; i suggest you to carry out the calculation again by plugging the right value in instead of the guessed 2

since pulley is guessed homogeneous , then the translating masses can be seen as rotating masses by multiplying them by 2

total rotating mass mr = mpu+(m1+m2)*2 = 1.9+6*2 = 13.9  kg

moment of inertia J = mr/2*(0.12/2)^2 = 13.9/2*36/10^4 = 0.0250 kg*m^2

motive torque Tm = g(4-2)*r = 9.807*2*0.06 = 1.177 N*m

accelerating torque Ta = Tm-Tf = 1.177-0.60 = 0.577 N*m

angular acceleration α = Ta/J = 0.577/0.0250 = 23.0 rad/sec^2

tangential acceleration a = α *r = 23.0*0.06 = 1.38 m/sec^2

called h the distance to the floor :

t = √ 2h/a

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