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Chapter 18: Problem 36

The assembly consists of a \(3-\mathrm{kg}\) pulley \(A\) and \(10-\mathrm{kg}\) pulley \(B\). If a \(2-\mathrm{kg}\) block is suspended from the cord, determine the block's speed after it descends \(0.5 \mathrm{~m}\) starting from rest. Neglect the mass of the cord and treat the pulleys as thin disks. No slipping occurs.

Short Answer

Expert verified
The speed of the block after it descends 0.5m starting from rest is 0.7 m/s.

Step by step solution

01

Calculating Moment of Inertia

First, the moment of inertia of the pulleys needs to be found. Since both are treated as thin disks, their moment of inertia is \(I = 0.5*m*r^2\). Here, we are not given the radii of the pulleys so we can assume them to be \(r_k\) for pulley \(k\). As such, the moment of inertia of pulley A is \(I_A = 0.5*m_A*r_A^2 = 1.5*r_A^2\) and for pulley B it is \(I_B = 0.5*m_B*r_B^2 = 5*r_B^2\). We are not given the radii, but in this case, it's not a problem, as \(r_A\) and \(r_B\) will cancel out in the next steps.
02

Conservation of Energy

At the start, the system is at rest, thus kinetic energy is zero and potential energy is \(mg*h = 2*9.81*0.5 = 9.81 J\). This energy transforms into the rotational energy of the pulleys and the kinetic energy of the block as it descends. So, we end up with initial potential energy equal to the final kinetic energy plus the total rotational energy, i.e., \(2*9.81*0.5 = 0.5*2*V^2 + 0.5*(1.5*r_A^2 + 5*r_B^2)(V/r)^2\).
03

Solve for V

From Step 2, we can simplify the equation to \(4.9 = V^2 + 3*r_A^2*V^2 + 10*r_B^2*V^2\). Because the string doesn't slip on the pulleys, the linear velocity of the string equals the tangential velocity of the pulley rims, i.e., \(V = r_A*w_A = r_B*w_B\) meaning \(r_A = r_B\), thus the equation simplifies to \(4.9 = 14*V^2\). Solving for V gives \(V = sqrt(4.9/14) = 0.7 m/s\).

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Most popular questions from this chapter

The double pulley consists of two parts that are attached to one another. It has a weight of \(50 \mathrm{lb}\) and a centroidal radius of gyration of \(k_{o}=0.6 \mathrm{ft}\) and is turning with an angular velocity of \(20 \mathrm{rad} / \mathrm{s}\) clockwise. Determine the angular velocity of the pulley at the instant the \(20-1 \mathrm{~b}\) weight moves \(2 \mathrm{ft}\) downward.C

The \(10-\) lb sphere starts from rest at \(\theta=0^{\circ}\) and rolls without slipping down the cylindrical surface which has a radius of \(10 \mathrm{ft}\). Determine the speed of the sphere's center of mass at the instant \(\theta=45^{\circ}\).

The two \(2-\mathrm{kg}\) gears \(A\) and \(B\) are attached to the ends of a \(3-\mathrm{kg}\) slender bar. The gears roll within the fixed ring gear \(C\), which lies in the horizontal plane. If a \(10-\mathrm{N} \cdot \mathrm{m}\) torque is applied to the center of the bar as shown, determine the number of revolutions the bar must rotate starting from rest in order for it to have an angular velocity of \(\omega_{A B}=20 \mathrm{rad} / \mathrm{s}\). For the calculation, assume the gears can be approximated by thin disks. What is the result if the gears lie in the vertical plane?

If the gear is released from rest, determine its angular velocity after its center of gravity \(O\) has descended a distance of \(4 \mathrm{ft}\). The gear has a weight of \(100 \mathrm{lb}\) and a radius of gyration about its center of gravity of \(k=0.75 \mathrm{ft}\).

The force of \(T=20 \mathrm{~N}\) is applied to the cord of negligible mass. Determine the angular velocity of the \(20-\mathrm{kg}\) wheel when it has rotated 4 revolutions starting from rest. The wheel has a radius of gyration of \(k_{O}=0.3 \mathrm{~m}\).
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