Equilibrium Applications of Newton's Laws of Rotation

[SUchica10]

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.)

 

[PhanthomJay]

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.

 

[kyle8921]

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