Equilibrium Applications of Newton's Laws of Rotation

In summary, the conversation discusses the leaning Tower of Pisa, its current displacement from the vertical, and the additional displacement needed to bring it to the verge of toppling. It also mentions the angle the tower will make with the vertical at that moment and the importance of finding the center of mass to determine the point at which the tower will become unstable.
  • #1
SUchica10
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Hi, I am stuck on this problem, any feedback would be greatly appreciated. Thank you...

The leaning Tower of Pisa is 55 m high and 7.0 m in diameter. The top of the tower is displaced 4.5 m from the vertical. Treating the tower as a uniform, circular cylinder, (a) what additional displacement, measured at the top, will bring the tower to the verge of toppling? (b) What angle with the vertical with the tower make at that moment? (The current rate of movement of the top is 1 mm/year.)
 
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  • #2
SUchica10 said:
Hi, I am stuck on this problem, any feedback would be greatly appreciated. Thank you...

The leaning Tower of Pisa is 55 m high and 7.0 m in diameter. The top of the tower is displaced 4.5 m from the vertical. Treating the tower as a uniform, circular cylinder, (a) what additional displacement, measured at the top, will bring the tower to the verge of toppling? (b) What angle with the vertical with the tower make at that moment? (The current rate of movement of the top is 1 mm/year.)
Find the c.m. of the tower. At what point will the c.m. be directly above the lower right corner of the tower (at the 7 m mark)? This is where it will become unstable due to the overturning torque from the tower's weight not being able to countered by the ground support force and torque that will no longer exist.
 
Last edited:
  • #3


Hello! I would be happy to provide some feedback on this problem.

To begin, let's first define some variables that we will use in our solution:

h = height of the tower (55 m)
d = diameter of the tower (7.0 m)
θ = angle between the tower and the vertical
x = additional displacement needed at the top to bring the tower to the verge of toppling
ω = angular velocity (given by the current rate of movement of the top, 1 mm/year)

Now, let's apply Newton's laws of rotation to this problem. Newton's first law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. In this case, the tower is not rotating, so we can ignore this law.

Newton's second law states that the torque (τ) acting on an object equals the moment of inertia (I) times the angular acceleration (α). In this case, the torque is caused by the gravitational force (mg) acting on the center of mass of the tower, which is located at the midpoint of its height. Therefore, we can write the equation: τ = Iα.

We know that the moment of inertia for a uniform, circular cylinder is given by I = (1/2)mr^2, where m is the mass of the cylinder and r is the radius. In this case, the mass of the tower is not given, but we can calculate it using its volume and density. The volume of a cylinder is given by V = πr^2h, and the density (ρ) is given by ρ = m/V. Therefore, the mass of the tower is m = ρV = ρπr^2h.

Now, let's plug in our values and solve for the torque:

τ = (mg)(h/2) = (ρπr^2h)(g)(h/2) = (ρπr^2h^2)(g/2)

Since we want to find the additional displacement needed at the top to bring the tower to the verge of toppling, we can set this torque equal to the torque caused by the additional displacement, which is given by τ = xmg. Therefore, we can write the equation:

xmg = (ρπr^2h^2)(g/2)

Solving for x, we get:

x = (ρπr^2
 

1. What is the concept of equilibrium in Newton's Laws of Rotation?

In Newton's Laws of Rotation, equilibrium refers to the state in which an object's rotational motion does not change. This means that the net torque acting on the object is equal to zero.

2. How is equilibrium achieved in rotational motion?

Equilibrium in rotational motion is achieved when the sum of all the torques acting on an object is equal to zero. This can be achieved by balancing the forces acting on the object or by adjusting the object's center of mass.

3. What are some real-life examples of equilibrium in rotational motion?

Some real-life examples of equilibrium in rotational motion include a spinning top, a bicycle wheel in motion, and a Ferris wheel. In all of these examples, the object's rotational motion remains constant due to the balanced torques acting on them.

4. How does the concept of equilibrium apply to objects with irregular shapes?

The concept of equilibrium applies to objects with irregular shapes in the same way as it does to objects with regular shapes. The key is to find the object's center of mass and ensure that the net torque acting on it is equal to zero.

5. Can an object be in both translational and rotational equilibrium at the same time?

Yes, an object can be in both translational and rotational equilibrium at the same time. This means that the object's motion remains constant both in terms of its position and its rotation.

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