What Is the Acceleration of the 8lb Block in a Yo-Yo Pulley System?

[lehel]

Homework Statement

a 16lb yo-yo is resting on a horizontal table and is free to roll. The dimensions are r = 4in and R = 12in and the radius of gyration about the CM is 6in. A cord is wrapped around the inner radius, the passes over a perfect pulley, and is attached to a 8lb block.

Find acceleration of 8lb block and the tension in the cord

(answers are 4.830ft/s^2 and 6.792lb)

Homework Equations

F = ma
torque = I*alpha = F*moment-arm
Iaxis = Icm + mh^2
I = (radius-gyration)R^2

The Attempt at a Solution

there are two accelerations a1 (weight) and a2 (yo-yo).

(16/32slugs)*a2 = T

I at table contact point = 90slug*in^2

net torqe at table contact point = 90slug*in^2 * alpha = T(12in - 4in)

T = 11.25*alpha and 0.5slugs * a2 = T

for weight, (8/32slugs)a1 = 8lb - T

so far, I have 4 unknowns and 3 equations so I can't solve for anything. What equation am I missing? i don't know how to find a relationship between alpha and a2 (i thought a2 = 12in*alpha but that doesn't work out right).

thanks in advance

 

[anathema]

for the help



Hello, I will try to help you with your problem. First, we need to determine the forces acting on the system. We have the weight of the yo-yo (16lb) and the weight of the block (8lb) acting downwards. The tension in the cord will be acting upwards. Since the yo-yo is resting on a horizontal table, we can assume that there is no friction.

Next, we can draw a free body diagram for the block and the yo-yo separately. For the block, we have its weight (8lb) acting downwards and the tension T in the cord acting upwards. The net force on the block will be equal to its mass (8/32 slugs) times its acceleration a1.

For the yo-yo, we have its weight (16lb) acting downwards and the tension T in the cord acting upwards. The net force on the yo-yo will be equal to its mass (16/32 slugs) times its acceleration a2.

Now, we can apply Newton's second law (F=ma) to both the block and the yo-yo. For the block, we have:

8lb - T = (8/32 slugs) * a1

For the yo-yo, we have:

T - 16lb = (16/32 slugs) * a2

Next, we can apply the equation for torque (torque = I*alpha) to the yo-yo. The torque acting on the yo-yo is due to the tension in the cord and is equal to T*(12in - 4in) = 8in*T. The moment of inertia of the yo-yo about its center of mass is given by I = (radius of gyration)^2 * mass = (6in)^2 * (16/32 slugs) = 3slugs*in^2. Thus, we have:

8in*T = 3slugs*in^2 * alpha

We can substitute this back into the equation for the yo-yo to get:

T - 16lb = (16/32 slugs) * (3slugs*in^2 * alpha)

T - 16lb = 1.5in * alpha

Now, we have two equations (one for the block and one for the yo-yo) and two unknowns (a1 and a2). We can solve these equations

 

[Moetasim]

To solve this problem, we can use the equations for rotational motion and Newton's second law. First, let's consider the motion of the yo-yo. The torque on the yo-yo is due to the tension in the cord and is given by:

τ = TR = T(R-r)

where T is the tension in the cord and R is the outer radius of the yo-yo. Using the equation τ = Iα, where I is the moment of inertia and α is the angular acceleration, we can set up the following equation:

T(R-r) = Iα

Since we know the radius of gyration, we can calculate the moment of inertia about the CM using the equation I = Icm + mh^2, where Icm is the moment of inertia about the CM and m is the mass of the yo-yo. Substituting in the values given in the problem, we get:

I = (90 slug*in^2) + (16/32 slug)(6in)^2 = 90 slug*in^2 + 9 slug*in^2 = 99 slug*in^2

Now, we can solve for α:

T(R-r) = Iα

T(12in-4in) = 99 slug*in^2 * α

T(8in) = 99 slug*in^2 * α

α = (T*8in)/(99 slug*in^2)

Next, let's consider the motion of the 8lb block. The only force acting on the block is its weight, which is given by:

F = ma

where m is the mass of the block and a is its acceleration. We can solve for a:

a = F/m = (8lb)/(8/32 slug) = 4 ft/s^2

Now, we can use the equation F = ma to solve for the tension in the cord:

T = ma = (8/32 slug) * (4 ft/s^2) = 1 slug*ft/s^2 = 6.792 lb

Finally, we can use the equation a = αr to solve for the acceleration of the block, where r is the radius of the yo-yo:

a = αr = [(T*8in)/(99 slug*in^2)]*4in = (T*32in^2)/(99 slug*in^2) = (6.