Physics+ Newton Laws, Problem 5: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton - Problem 5

Problem A

A point mass M is uniformly rotating within a plane with XY Cartesian coordinates on radius R and angular speed of rotation ω on a weightless unstretchable thread around a point that is the center of coordinates on this plane.
Determine the force F with which a thread holds this point mass on its fixed orbit (centripetal force) and express this force in Cartesian coordinates [F_x,F_y].
What is the direction and magnitude of this force?

Solution A

Position of a rotating point mass, as a function of time t, is a vector r(t) with the following coordinates
X(t) = R·cos(ω·t)
Y(t) = R·sin(ω·t)

Coordinates of the velocity vector V=[V_x,V_y] are, by definition, derivatives of coordinates of position
V_x(t) = −R·ω·sin(ω·t)
V_y(t) = R·ω·cos(ω·t)

Vector of velocity is perpendicular to a position vector. To confirm it, let's form a scalar product of these two vectors and check if it's equal to zero.
X(t)·V_x(t) + Y(t)·V_y(t) =
= −R²·ω·cos(ω·t)·sin(ω·t) +
+ R²·ω·sin(ω·)·cos(ω·t) = 0

Coordinates of the acceleration vector a=[a_x,a_y] are, by definition, derivatives of coordinates of the velocity vector
a_x(t) = −R·ω²·cos(ω·t)
a_y(t) = −R·ω²·sin(ω·t)

Since the centripetal force and acceleration are related, according to the Newton's Second Law, as F=M·a,
F_x(t) = M·a_x = −M·R·ω²·cos(ω·t)
F_y(t) = M·a_x = −M·R·ω²·sin(ω·t)

As we see, the vector of force is collinear and opposite in direction to the vector of position, that is, it's directed towards the center of rotation.

At the same time, the vector of force, being collinear to the vector of position, is perpendicular to the vector of velocity.

The magnitude of the centripetal force is
|F(t)| = M·R·ω²


Problem B

In the context of the Problem A above, consider that the plane of rotation is made of some frictionless material and positioned perpendicularly to the force of gravity. There is a hole in the center of rotation and a thread that holds the rotating point mass M goes down through that hole and is held by hand in a fixed position.
The radius of rotation is the same as above R and the angular speed is ω.
What mass m₀ should be attached to the bottom of a thread to replace the holding hand to maintain original radius and angular speed of rotation?
How the rotation will change (in terms of radius and angular speed) if we add another mass m₁ to the mass m₀ needed to maintain the original rotation?

Solution B

The results of the Problem A above regarding the direction and magnitude of the centripetal force F are:
1. The centripetal force is directed towards the center of the rotation, opposite to a position vector of the rotating object;
2. The magnitude of this force is |F(t)| = M·R·ω²

The condition of the Problem B is that the source of the centripetal force holding an object on a fixed orbit of rotation is the holding hand. If we replace it with the gravitation force of an object of mass m₀, the magnitude of the gravitational force must be equal to the one calculated above:
m₀·g = M·R·ω²
Therefore,
m₀ = M·R·ω²/g

Adding another mass m₁ to the above will change the picture. Both radius and angular frequency of rotation will change to correspond to a changed magnitude in the centripetal force.
At the same time, all forces acting on a rotating object are central, and that is a sufficient condition for the Law of Conservation of Angular momentum.
The above is sufficient to calculate a new radius and angular speed of a rotating object.

New magnitude of a centripetal force is (m₀+m₁)·g.
Therefore,
(m₀+m₁)·g = M·R_new·ω²_new
This is one equation with two variables to find:
R_new and ω_new.

Taking into account the Law of Conservation of the Angular Momentum will produce another equation, and that should be sufficient to find both unknown variables.

The magnitude of the original angular momentum of an object rotating on radius R with the angular speed ω is
L = |L| = |r⨯ p|
where
r is a position vector of a rotating object and
p is a linear momentum of a rotating object.

Linear speed v of an object rotating on a radius R with an angular speed ω is R·ω.
Since r is a radial vector and p=M·v is tangential to an orbit, these two vectors are perpendicular to each other, and the magnitude of their vector product is a product of their magnitudes.
Therefore,
L = M·R·v = M·R²·ω

After we changed the centripetal force the angular momentum must be the same. Therefore,
L = M·R²·ω =
= L_new = M·R²_new·ω_new
or
R²·ω = R²_new·ω_new

This produces the second equation we need to determine new radius and angular velocity. Hence, the system of two equations with two unknows is
(m₀+m₁)·g = M·R_new·ω²_new
R²·ω = R²_new·ω_new

Resolving the second equation for ω_new and substituting it into the first produces
ω_new = R²·ω/R²_new
(m₀+m₁)·g =
= M·R_new·(R²·ω/R²_new)² =
= M·R⁴·ω²/R³_new

From the above follows the value for a new radius of rotation
R³_new = M·R⁴·ω²/[(m₀+m₁)·g]
R_new =
= {M·R⁴·ω²/[(m₀+m₁)·g]}^⅓
The new angular speed can be derived from the above and the condition ω_new=R²·ω/R²_new
ω_new = R²·ω·
·{(m₀+m₁)·g/[M·R⁴·ω²]}^⅓

This rather complex formula needs to be verified somehow.
Let's consider a simple case of a radius shortened by half and check what additional mass m₁ is needed.
In a formula for R³_new above we will set R_new=½R.
(½R)³ = M·R⁴·ω²/[(m₀+m₁)·g]
from which follows
1 = 8M·R·ω²/[(m₀+m₁)·g]
(m₀+m₁)·g = 8M·R·ω² =
= 8m₀·g
Therefore,
m₁ = 7m₀.
In other words, to decrease the radius by half we have to add 7 times the mass that used to hold a rotating object on its original distance from a center of rotation, that is we have to increase the mass by factor 8.

As we know (see Problem A above), the magnitude of the centripetal force is
|F(t)| = M·R·ω²
Also known is that, when the radius of rotation shortens by half, the angular velocity increases by factor 4.
From these two statements follows that a centripetal force in this case increases by a factor of 8, which corresponds to our calculations above.