A block with mass m = 5.00 kg slides down a surface inclined 36.9 to the horizon

Question

A block with mass m = 5.00 kg slides down a surface inclined 36.9 to the horizontal (the figure (Figure 1) ). The coefficient of kinetic friction is 0.27. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 16.0 kg and moment of inertia 0.500 kgm2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.250 m from that axis. What is the acceleration of the block down the plane? What is the tension in the string?

Explanation / Answer

First take just the block. The sum of forces = ma, right.

Call F the friction force and Wp, the component of the weight parallel to the slope.

ma = Wp - T - F, right?

Well F = u * m * g cos angle, and Wp = m * g * sin angle, so you can calculate those out if you wish.

F = 0.27 * 5 kg * 9.81 m/s^2 cos 36.9

F = 10.59 N

Wp = 5 kg * 9,.81 m/s^2 * sin 36.9

Wp = 29.5 N

Our other equation is sum of torques = moment of inertia (I) times angular acceleration (call it alpha)

so the only torque is the tension of the string times the radius (R)

T * R = I * alpha

And one more equation, alpha = a/R

solve for T

T = I * a /R^2

Now plug that into the other equation

ma = Wp - F - (I * a/R^2)

Now just solve for acceleration.

ma + (I * a/R^2) = Wp - F

a (m + (I/R^2) = Wp - F

So the m + (I/R^2) part is 5 kg + (0.500 kg m^2 / (0.250 m)^2) = 13 kg

a = (29.5 N - 10.59 N)/13 kg

a = 1.45 m/s^2 (ans)

using T = I * a / R^2 to solve for T

T = {0.500 kg m^2 * 1.45 m/s^2} / (0.250 m)^2

T = 11.64 N (ans)

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