The 4.00-kg block in the figure is attached to a vertical rod by means of two strings. When the system rotates about the axis of the rod, the strings are extended towards the rod and each have a length of 1.25. The length from the upper string and lower string on the rod is 2. So it forms a triangle with sides 2, 1.25, 1.25; with the block between the 1.25 sides. The tension in the upper string is 85.0 N . Tension of the lower cord is 36 N.

a) How many revolutions per minute does the system make?

b)Find the number of revolutions per minute at which the lower cord just goes slack.

a) Well, this block is spinning around like a breakdancer on the dance floor. To find the number of revolutions per minute, we can use a little geometry and physics. The length of the upper string is 1.25, and since it's spinning in a circle, that means the circumference of that circle is 2π times the length of the string. That would be 2π * 1.25 = 7.85 meters. Now, we need to find the distance the block travels in one minute. The distance traveled in one revolution is equal to the circumference, which is 7.85 meters. So in one minute, the block would cover 7.85 meters * the number of revolutions. To find the number of revolutions per minute, we divide the distance by the length of the string. So it's (7.85 meters)/(1.25 meters) = 6.28 revolutions per minute. Keep on spinning, block!

b) Ah, the lower cord is about to get all loosey-goosey, huh? Well, to find the number of revolutions per minute at which the lower cord just goes slack, we need to figure out the tension in the lower cord when it goes slack. When the lower cord goes slack, the tension is zero. So we need to figure out at what speed the centripetal force equals zero, thus making the tension zero. The centripetal force is given by the tension in the lower cord, so when the tension is zero, the centripetal force is zero. We can use the formula for centripetal force, which is F = mω²r, where m is the mass of the block, ω is the angular velocity (in radians per second), and r is the radius of the circular path. We already know the mass of the block (4 kg) and the radius of the circular path (1.25 m). So we need to solve for ω when F = 0. That means 0 = mω²r. Rearranging that equation, we get ω = 0 (since m and r are both positive values). So the angular velocity ω is zero when the tension in the lower cord is zero. Therefore, the lower cord goes slack when there is no rotation, meaning zero revolutions per minute. Keep that lower cord taut, my friend!

To solve this problem, you can use the concept of centripetal force. The tension in the strings provides the centripetal force required to keep the block rotating in a circle.

a) To find the number of revolutions per minute the system makes, you can use the formula for centripetal force:

F = m * (v^2 / r)

Where F is the centripetal force, m is the mass of the block, v is the velocity, and r is the radius.

In this case, the tension in the upper string provides the centripetal force. The tension in the upper string is given as 85.0 N. The length of the upper string is 1.25 m. The radius, r, can be calculated by subtracting the length of the upper string from the length of the triangle side (2 m):

r = 2 - 1.25
r = 0.75 m

Now, we can calculate the velocity using the centripetal force formula:

F = m * (v^2 / r)
85.0 N = 4.00 kg * (v^2 / 0.75 m)
85.0 N * 0.75 m = 4.00 kg * v^2
63.75 N*m = 4.00 kg * v^2
v^2 = 63.75 N*m / 4.00 kg
v^2 = 15.9375 m^2/s^2
v = sqrt(15.9375 m^2/s^2)
v ≈ 3.999 m/s

Now, to find the number of revolutions per minute (rpm), you can use the formula:

v = (2πr * n) / 60

Where n is the number of revolutions per minute.

Rearranging the formula, we can solve for n:

n = (v * 60) / (2πr)
n = (3.999 m/s * 60 s/min) / (2π * 0.75 m)
n ≈ 31.9 rpm

Therefore, the system makes approximately 31.9 revolutions per minute.

b) To find the number of revolutions per minute at which the lower cord just goes slack, we need to find the minimum centripetal force required to keep the block rotating. This minimum centripetal force corresponds to the tension in the lower cord.

The tension in the lower cord is given as 36 N. Using the same formula for centripetal force as before, we can calculate the velocity required for the lower cord tension:

36 N = 4.00 kg * (v^2 / 0.75 m)
36 N * 0.75 m = 4.00 kg * v^2
27 N*m = 4.00 kg * v^2
v^2 = 27 N*m / 4.00 kg
v^2 = 6.75 m^2/s^2
v = sqrt(6.75 m^2/s^2)
v ≈ 2.598 m/s

Now, we can use the formula to find the number of revolutions per minute:

n = (v * 60) / (2πr)
n = (2.598 m/s * 60 s/min) / (2π * 0.75 m)
n ≈ 20.8 rpm

Therefore, the lower cord just goes slack at approximately 20.8 revolutions per minute.

To answer both parts of the question, we need to first understand the forces acting on the block and the rotational motion of the system.

Let's start by analyzing the forces on the block when the system is rotating. There are two tension forces acting on the block: one from the upper string and one from the lower string. The gravitational force acting on the block will also come into play.

a) To find the number of revolutions per minute (rpm), we need to use the relationship between the centripetal force and rotational motion.

The centripetal force acting on an object rotating in a circle is given by the equation:

F = m * r * ω^2

where F is the centripetal force, m is the mass of the object, r is the radius of the circle (in this case, the distance from the center of rotation to the block), and ω is the angular velocity in radians per second.

In our case, the centripetal force on the block is provided by the tension in the upper string, which is given as 85.0 N.

Substituting the values into the equation, we have:

85.0 = 4.00 * r * ω^2

Now, let's determine the radius of the circular motion. The length of the upper string is given as 1.25 units, and the block is located between two equal sides of length 1.25 units. Therefore, the radius r of the circular motion is half of the length of the upper string:

r = 0.5 * 1.25
r = 0.625 units

Substituting the value of r into the equation, we have:

85.0 = 4.00 * 0.625 * ω^2

Now, solving for ω, we get:

ω^2 = 85.0 / (4.00 * 0.625)
ω^2 = 34 / 5
ω^2 = 6.8

Taking the square root of both sides of the equation, we find:

ω = √(6.8)
ω ≈ 2.61 radians/second

Finally, to find the number of revolutions per minute, we need to convert ω into units of rpm. There are 2π radians in one revolution, and 60 seconds in one minute:

rpm = (2.61 radians/second) * (60 seconds/1 minute) * (1 revolution/2π radians)

Calculating this expression, we find:

rpm ≈ 24.88 revolutions/minute

So, the system makes approximately 24.88 revolutions per minute.

b) Now let's determine the number of revolutions per minute at which the lower cord just goes slack. When the lower cord goes slack, the tension in the lower string becomes zero.

To find the lower cord's slack point, we need to analyze the forces acting on the block. When the tension in the lower string is zero, the only force remaining is the gravitational force acting on the block.

The gravitational force acting on the block is given by the equation:

F_grav = m * g

where F_grav is the force due to gravity, m is the mass of the object, and g is the acceleration due to gravity.

In our case, the force due to gravity is given as:

F_grav = 4.00 kg * 9.8 m/s^2
F_grav = 39.2 N

Since the lower string goes slack, the force due to gravity provides the necessary centripetal force for the block to stay in circular motion. Therefore, we equate the force due to gravity to the centripetal force:

F_grav = m * r * ω^2

Substituting the known values, we have:

39.2 = 4.00 * 0.625 * ω^2

Now, solving for ω, we get:

ω^2 = 39.2 / (4.00 * 0.625)
ω^2 = 15.68 / 2.5
ω^2 = 6.27

Taking the square root of both sides, we find:

ω = √(6.27)
ω ≈ 2.50 radians/second

To find the number of revolutions per minute, we use the same conversion as before:

rpm = (2.50 radians/second) * (60 seconds/1 minute) * (1 revolution/2π radians)

Calculating this expression, we find:

rpm ≈ 23.80 revolutions/minute

Therefore, the lower cord just goes slack at approximately 23.80 revolutions per minute.

For part a)

T1 = Upper cord
T2 = Lower cord
theta = angle between vertical rod and upper cord = arcos(1/1.25)
R = radius of circular motion = sqrt((1.25^2) + 1^2)
M = mass of object

T1sin(theta)+T2sin(theta) = (MV^2)/R

Solve for V

Turn your V (should be in m/s) into revs/min.

rev = circumference = 2Rpi
m/s * 1 rev/(2Rpi)m * 60s/1min = rev/min

V in rev/min = (60V)/(2Rpi)

Haven't figured out part b) yet