Review. An object with a mass of m 5 5.10 kg is

attached to the free end of a light string wrapped around
a reel of radius R 5 0.250 m and mass M 5 3.00 kg. The
reel is a solid disk, free to rotate in a vertical plane about
the horizontal axis passing through its center as shown in
Figure P10.51. The suspended object is released from rest
6.00 m above the floor. Determine (a) the tension in the
string, (b) the acceleration of the object, and (c) the speed
with which the object hits the floor. (d) Verify your answer
to part (c) by using the isolated system (energy) model.

To solve this problem, we can use the principles of rotational dynamics and conservation of energy.

(a) Tension in the string:
The tension in the string can be determined by considering the rotational equilibrium of the reel. Since the reel is rotating but not translating, the net torque acting on it must be zero.
The torque due to the weight of the object is given by T1 = m * g * R, where m is the mass of the object, g is the acceleration due to gravity, and R is the radius of the reel.

(b) Acceleration of the object:
The acceleration of the object can be determined using Newton's second law of motion. The net force acting on the object is the tension in the string minus the weight of the object:
T - m * g = m * a,
where T is the tension in the string and a is the acceleration of the object.

(c) Speed with which the object hits the floor:
To find the speed with which the object hits the floor, we can use the principle of conservation of energy. The initial gravitational potential energy of the object will be converted into kinetic energy as it falls.

(d) Verification using the isolated system (energy) model:
Using the principle of conservation of energy, we can compare the change in potential energy with the change in kinetic energy to verify our answer to part (c).

Let's calculate the values step-by-step.

(a) Tension in the string:
T1 = m * g * R,
T1 = (5.10 kg) * (9.8 m/s^2) * (0.250 m),
T1 = 12.57 N.

(b) Acceleration of the object:
T - m * g = m * a,
T - (5.10 kg) * (9.8 m/s^2) = (5.10 kg) * a,
12.57 N - 50.0 N = (5.10 kg) * a,
-37.43 N = (5.10 kg) * a,
a = -7.34 m/s^2 (negative sign indicates downward motion).

(c) Speed with which the object hits the floor:
Using the equation of motion, v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (0 m/s), a is the acceleration, and s is the distance travelled (6.00 m).
0 = 0 + 2 * (-7.34 m/s^2) * 6.00 m,
0 = -87.84 m^2/s^2,
Taking the square root of both sides, we get
v = √(87.84 m^2/s^2),
v = 9.37 m/s.

(d) Verification:
Initial potential energy = m * g * h,
m * g * h = (5.10 kg) * (9.8 m/s^2) * (6.00 m),
m * g * h = 299.88 J.

Final kinetic energy = 0.5 * m * v^2,
0.5 * (5.10 kg) * (9.37 m/s)^2,
Final kinetic energy = 221.93 J.

The change in energy = Final kinetic energy - Initial potential energy,
Change in energy = 221.93 J - 299.88 J,
Change in energy = -77.95 J.

The negative sign indicates that energy is lost, which is expected due to dissipative forces like air resistance.

Therefore, our answer in part (c) is verified using the isolated system (energy) model.

To solve this problem, we need to analyze the forces acting on the system and apply the principles of Newton's laws of motion and conservation of energy.

First, let's consider the forces acting on the object hanging from the string. We have the weight of the object (mg) acting downwards, and the tension (T) in the string acting upwards.

(a) To find the tension in the string:
Using Newton's second law, we know that the net force acting on the object is equal to its mass multiplied by its acceleration. Since the object is released from rest, its initial velocity is zero, and the only force acting on it is its weight. Therefore, we can write:

mg = ma

Now, we can rearrange the equation to solve for the tension T:

T = mg

Substituting the given values, T = (5.10 kg)(9.8 m/s^2) = 49.98 N (approximately 50 N)

(b) To find the acceleration of the object:
Using the same equation as above (mg = ma), we can rearrange it to solve for the acceleration a:

a = g

Substituting the value of acceleration due to gravity (g = 9.8 m/s^2), the acceleration of the object is 9.8 m/s^2.

(c) To find the speed with which the object hits the floor:
We can use the kinematic equation that relates the final velocity (v), initial velocity (u), acceleration (a), and displacement (s):

v^2 = u^2 + 2as

In this case, the initial velocity (u) is zero, as the object is released from rest. The displacement (s) is given as 6.00 m. Solving for v, we get:

v^2 = 0^2 + 2(9.8 m/s^2)(6.00 m)

v^2 = 2(9.8 m/s^2)(6.00 m) = 117.6 m^2/s^2

Taking the square root of both sides:

v ≈ √117.6 m^2/s^2 ≈ 10.85 m/s

Therefore, the speed with which the object hits the floor is approximately 10.85 m/s.

(d) To verify the answer using the isolated system (energy) model:
In an isolated system, the total mechanical energy remains constant. At the top, the object has potential energy (mgh), and at the bottom, it has kinetic energy (0.5mv^2). Thus, we can equate these energies:

mgh = 0.5mv^2

Simplifying:

gh = 0.5v^2

Plugging in the given values, we get:

9.8 m/s^2 * 6.00 m = 0.5 * v^2

58.8 m^2/s^2 = 0.5 * v^2

117.6 m^2/s^2 = v^2

Taking the square root of both sides:

v ≈ √117.6 m^2/s^2 ≈ 10.85 m/s

As you can see, the result matches the previous calculation.

Thus, the tension in the string is 50 N, the acceleration of the object is 9.8 m/s^2, and the speed with which the object hits the floor is approximately 10.85 m/s.

wr2

Look at energy.

Intial gravational energy=mass*g*h

rotaltional energy=1/2 I w^2=1/2 I(v/r)^2
(look up the moment of inertia for a solid disk I)

rotational energy+1/2 mv^2=Initial GPE
solve for v when it hits the floor (c) this will verify the following.

a) net force=torque=momentI*angularacceleration
= I*tangential acceleration/r=I*a/r

and net force=mg-ma

mg-ma=I a/r solve for a, (b), then solve for tension (mg-ma)

Now
a)