Question

An object with a mass of m = 5.15 kg is attached to the free endof a light string wrapped around a reel of radius R = 0.270 m andmass of M = 3.00 kg. The reel is a solid disk, free to rotate in avevertical plane about the horizontal axis passing through its center as shown in the figure below. The suspended object is released from rest \( 5.80 \mathrm{~m} \) above the floor. (a) Determine the t

Answer 1

To determine the speed of the center of the disk when the stick is vertical, we can use the principle of conservation of angular momentum. Once we have obtained the final angular velocity, we can calculate the speed of the center of the disk using the formula v = Rdisk · τfinal.

Step-by-step explanation:

To determine the speed of the center of the disk when the stick is vertical, we can use the principle of conservation of angular momentum. When the stick is horizontal, it has an initial angular momentum of zero. As the stick rotates and becomes vertical, the total angular momentum of the system (disk + stick) must be conserved. The moment of inertia of the system is given by I = Idisk + Istick, where Idisk is the moment of inertia of the disk and Istick is the moment of inertia of the stick.

Using the conservation of angular momentum, we have:

Linitial = Lfinal

0 = Iinitial · τinitial = Ifinal · τfinal

Since the moment of inertia of the system is constant, we can rewrite the equation as:

τfinal = τinitial · (Iinitial / Ifinal)

Substituting the expression for Initial and Ifinal:

τfinal = τinitial · [(Mdisk · Rdisk2 + Mstick · Lstick2) / (Mdisk · Rdisk2 + Mstick · Rstick2)]

where Mdisk is the mass of the disk, Mstick is the mass of the stick, Rdisk is the radius of the disk, Rstick is the radius of the stick, and Lstick is the length of the stick.

Once we have obtained the final angular velocity, we can calculate the speed of the center of the disk using the formula v = Rdisk · τfinal.

Answer 2

It takes approximately 1.11 seconds for the object to reach the ground.

How to find time?

Gravity (mg) acting downwards on the object

Tension in the string (T)

Moment of inertia (I) of the reel

Apply Newton's second law:

Fnet = ma

where

m = mass of the object and

a = acceleration

Resolve forces into components:

Tension (T) balances the weight of the object (mg)

Torque due to tension (Tr = RT) is equal to the net torque acting on the reel, which causes its angular acceleration (α)

Write the equations of motion:

T = mg (force balance on the object)

Iα = Tr

= RT (torque balance on the reel)

Relate the linear and angular accelerations:

a = Rα (the linear acceleration of the object is related to the angular acceleration of the reel by the radius)

From the first equation, we have T = mg. Substitute this into the second equation:

Iα = Rmg

Substitute a = Rα into the first equation:

mg = m(Rα)

= mR²α

Solve for the angular acceleration α:

α = g / R

The total mass of the system is M + m. Use this to calculate the moment of inertia of the reel:

I = 1/2 × M × R²

Substitute all the values into the torque equation:

1/2 × M × R² × (g / R) = Rmg

Cancel out common factors and solve for g:

g = 2 × M / (M + m)

Substitute this value of g back into the equation for α:

α = 2 × M × g / (M + m) × R

Finally, use the equation a = Rα to solve for the time (t) it takes for the object to reach the ground:

t = √(2h / a)

where h = initial height of the object (5.80 m).

Calculate the time:

t = √(2 × 5.80 m / (2 × 3.00 kg / (3.00 kg + 5.15 kg) × 0.270 m))

t = 1.11 s

Therefore, it takes approximately 1.11 seconds for the object to reach the ground.