*Notes to a video lecture on http://www.unizor.com*

__Problems on__

Rotational Dynamics

Rotational Dynamics

*Problem 1*

Consider a device that consists of a wheel of radius

**and mass**

*r***, freely rotating on some fixed axis, a thread rolled around it a few times and a point-object of mass**

*M***hanging off that thread.**

*m*Assume that all mass of a wheel is concentrated in its rim, while spokes are weightless.

There is no friction in the rotating wheel.

A thread is weightless and unstrechable.

The force of gravity pulls the object down with free fall acceleration

**.**

*g*What will be the acceleration of an object?

*Solution*

The weight of a wheel is balanced by the reaction of fixed axis and, therefore, can be ignored.

The only force acting tangentially on a wheel is the tension of a tread

**trying to rotate it.**

*T*Therefore, we can equate its momentum

**(**

*τ=T·r**torque*) to a product of its

*moment of inertia*

**and**

*I**angular acceleration*

**:**

*α*

*τ = T·r = I·α*The fact that all mass of a wheel is concentrated in its rim allows us

to easily calculate its moment of inertia. We can imagine a rim

consisting of a large number

**of point-objects of mass**

*N***each. The moment of inertia of each is**

*M/N***and combined moment of inertia of a wheel is**

*(M/N)·r²***. If the mass is not concentrated in the rim, analogous logic would lead us to more involved calculations of inertial mass.**

*N·(M/N)·r²=M·r²*Considering,

**and**

*I=M·r²***(where**

*r·α=a***is a**

*a**linear acceleration*of a thread) we can write the following equation:

*T·r = M·r²·a/r = M·r·a*Radius cancels out and we get

*T = M·a*Notice that this is exactly the same equation as the Newton's Second Law, as if an object of mass

**is pulled along a straight line by a force**

*M***with linear acceleration**

*T***.**

*a*Another important detail is that this formula is independent of a radius

of a wheel - a direct consequence of the fact that the mass of a wheel

is concentrated in its rim.

Analyzing the movement of an object hanging on a thread, we conclude that it moves with linear acceleration

**down by the forces of gravity (directed down) and tension of a thread (directed up). Therefore, using the Newton's Second Law,**

*a*

*m·g − T = m·a*Now we have a system of two equations with two unknowns

**and**

*T***, which is easy to solve.**

*a*Substitute

**from the first equation to the second:**

*T*

*m·g − M·a = m·a*

*m·g = (M + m) · a*

*a = m·g / (M + m)**Problem 2*

Calculate a moment of inertia of disk of mass

**and radius**

*M***rotating around an axis going through its center perpendicularly to its surface.**

*R**Solution*

Divide a disk into a set of concentric rings of infinitesimal width

*d*. Let the inner radius of a particular ring be

**r****and it outer radius be**

*r*

*r+**d*.

**r**Moment of inertia of each ring is

*I(r) = m·r²*where

**is its mass (see the previous Problem 1 for explanation).**

*m*Mass of a ring

**is a mass of a disk**

*m***multiplied by a ratio of a ring's area**

*M*

*2πr·**d*to the area of an entire disk

**r****.**

*πR²*So, the moment of inertia of our ring of radius

**and infinitesimal width**

*r**d*is

**r****[**

*I(r) =*

*M·2πr·**d*]

**r/(πR²)**

*·r²*Integrating this from

**to**

*r=0***, we get**

*r=R***[**

*I*

= ∫_{disk}(M, R) == ∫

*M·2πr·**d*]

**r/(πR**^{2})

*·r*

= (2M/R^{2}== (2M/R

^{2})∫r^{3}*d*

**r =**

= (2M/R

= M·R= (2M/R

^{2})·(r^{4}/4) == M·R

^{2}/2*Problem 3*

Consider Problem 1, but change a wheel with all mass concentrated in the

rim with a wheel with mass evenly distributed inside the circumference,

making it a disk.

*Solution*

We follow the same logic as in Problem 1.

The weight of a wheel is balanced by the reaction of fixed axis and, therefore, can be ignored.

The only force acting tangentially on a wheel is the tension of a tread

**trying to rotate it.**

*T*Therefore, we can equate its momentum

**(**

*τ=T·r**torque*) to a product of its

*moment of inertia*

**and**

*I**angular acceleration*

**:**

*α*

*τ = T·r = I·α*The fact that all mass of a wheel is evenly distributed within its

circumference allows us to easily calculate its moment of inertia using

the Problem 2 above:

*I=M·r²/2*Linear acceleration

**and angular acceleration**

*a***are related:**

*α*

*r·α=a*Therefore,

**[**

*T·r =***]**

*M·r²/2*

*·(a/r) = M·r·a/2*Radius cancels out and we get

*T = M·a/2*An important detail is that this formula is independent of a radius of a wheel.

Analyzing the movement of an object hanging on a thread, we conclude that it moves with linear acceleration

**down by the forces of gravity (directed down) and tension of a thread (directed up). Therefore, using the Newton's Second Law,**

*a*

*m·g − T = m·a*Now we have a system of two equations with two unknowns

**and**

*T***, which is easy to solve.**

*a*Substitute

**from the first equation to the second:**

*T*

*m·g − M·a/2 = m·a*

*m·g = (m + M/2) · a*

*a = m·g / (m + M/2)*If we compare this formula with the one in Problem 1, we see that final acceleration is greater because denominator is smaller.

So, the disk in this problem will rotate faster than a wheel with empty middle part from Problem 1.

*Problem 4*

Let's rotate a small ball of mass

**within a horizontal plane on a thread of length**

*M***. The thread will make certain angle**

*D***with the horizon. Experiments show that the angle will be smaller if a ball rotates faster.**

*φ*Determine the relationship between angular speed of rotation

**, mass of a ball**

*ω***, length of a thread**

*M***and angle**

*D***, ignoring friction and air resistance.**

*φ**Solution*

The tension of a thread

**keeps a ball on its orbit. Radius of an orbit is**

*T***.**

*R=D·cos(φ)*The tension of a thread

**serves dual purpose - its vertical component**

*T***acts against gravity**

*T·sin(φ)***, its horizontal component**

*M·g***acts as a centripetal force and should be equal to**

*T·cos(φ)***, where**

*M·V*^{2}/R**is a linear speed of a ball, which is equal, in turn, to**

*V***, where**

*R·ω***is angular speed of rotation of a ball.**

*ω*So, we have the following equations:

*T·sin(φ) = M·g*

*T·cos(φ) = M·(V*^{2})/R = M·R·ω^{2}= M·D·cos(φ)·ω^{2}The second equation is simplified to

*T = M·D·ω*^{2}Substituting it to the first equation,

*D·ω*^{2}·sin(φ) = gTherefore,

*sin(φ) = g/[D·ω*^{2}]For a given angular speed we can find an angle of a thread to horizon

**. It in inversely proportional to a length of a thread**

*φ***and to a square of angular speed**

*D***, which seems to be reasonable.**

*ω*Interestingly, this angle is independent of the mass

**.**

*M*