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Angular Velocity of Rod-Object System with Sliding Object

Created by @nathanedwards
 at November 1st 2023, 12:36:57 am.
AP_Physics_1-Questions
#mechanics
#physics
#rotational-motion

Question:

A uniform thin rod of length L and mass M is rotating about an axis perpendicular to its length and passing through its center. The moment of inertia of the rod about this axis is given by the expression:

I=112ML2I = \frac{1}{12} ML^2

An object of mass m is attached to one end of the rod as shown in the figure below. The system is initially at rest. The object starts sliding down the rod due to gravitational force, while the rod is allowed to rotate freely. Neglecting any frictional effects, determine the angular velocity of the rod-object system when the object reaches the end of the rod in terms of m, M, L, and g.

Rod with object

Answer:

To solve this problem, we can apply the principle of conservation of mechanical energy.

When the object reaches the end of the rod (position B), it has converted its initial gravitational potential energy into both kinetic energy and rotational kinetic energy.

The gravitational potential energy (GPE) at the start (position A) is given by:

GPEstart=mghGPE_{\text{start}} = mgh

where h is the vertical distance between the center of mass of the rod-object system and the height at position A.

Since we know that the center of mass of the rod is at its midpoint, h can be calculated as h=L2h = \frac{L}{2}.

Thus,

GPEstart=mgh=mg⋅L2GPE_{\text{start}} = mgh = mg \cdot \frac{L}{2}

At position B, the system has rotational kinetic energy (due to the rod) and translational kinetic energy (due to the object).

The translational kinetic energy (TKE) of the object at position B is given by:

TKEobject=12mv2TKE_{\text{object}} = \frac{1}{2} m v^2

where v is the velocity of the object at position B.

The rotational kinetic energy (RKE) of the rod-object system at position B is given by:

RKEsystem=12Isystemω2RKE_{\text{system}} = \frac{1}{2} I_{\text{system}} \omega^2

where IsystemI_{\text{system}} is the moment of inertia of the rod-object system and ω\omega is its angular velocity.

Since the rod-object system rotates about an axis through its center perpendicular to its length, the moment of inertia IsystemI_{\text{system}} can be calculated using the expression given in the question: Isystem=112ML2I_{\text{system}} = \frac{1}{12} ML^2.

The total mechanical energy (ME) of the system is conserved:

MEstart=MEendME_{\text{start}} = ME_{\text{end}}
GPEstart=TKEobject at B+RKEsystem at BGPE_{\text{start}} = TKE_{\text{object at B}} + RKE_{\text{system at B}}

Substituting the relevant equations:

mg⋅L2=12mv2+12Isystemω2mg \cdot \frac{L}{2} = \frac{1}{2} m v^2 + \frac{1}{2} I_{\text{system}} \omega^2

Now, let's find an expression for v in terms of angular velocity ω\omega.

v is the tangential velocity of the object at position B due to the rotation of the rod. The distance from the axis of rotation (center of the rod) to the object is equal to L2\frac{L}{2}. Therefore, the tangential velocity is given by:

v=ω⋅L2v = \omega \cdot \frac{L}{2}

Substituting this expression into the equation:

mg⋅L2=12m(ω⋅L2)2+12Isystemω2mg \cdot \frac{L}{2} = \frac{1}{2} m \left(\omega \cdot \frac{L}{2}\right)^2 + \frac{1}{2} I_{\text{system}} \omega^2

Simplifying the equation:

mg⋅L2=12m⋅ω2⋅(L2)2+12(112ML2)ω2mg \cdot \frac{L}{2} = \frac{1}{2} m \cdot \omega^2 \cdot \left(\frac{L}{2}\right)^2 + \frac{1}{2} \left(\frac{1}{12} ML^2\right) \omega^2

Now, let's solve for ω\omega by isolating it in the equation:

mg⋅L2=12m⋅ω2⋅L24+12⋅112⋅M⋅L2⋅ω2mg \cdot \frac{L}{2} = \frac{1}{2} m \cdot \omega^2 \cdot \frac{L^2}{4} + \frac{1}{2} \cdot \frac{1}{12} \cdot M \cdot L^2 \cdot \omega^2

Multiplying through by 4 to clear the fraction, and multiplying through by 12 to simplify:

2mgL=m⋅ω2⋅L2+112⋅4⋅M⋅L2⋅ω22mgL = m \cdot \omega^2 \cdot L^2 + \frac{1}{12} \cdot 4 \cdot M \cdot L^2 \cdot \omega^2
2mg=m⋅ω2⋅L+13⋅M⋅L2⋅ω22mg = m \cdot \omega^2 \cdot L + \frac{1}{3} \cdot M \cdot L^2 \cdot \omega^2

Factorizing out ω2\omega^2:

2mg=ω2⋅(m⋅L+13⋅M⋅L2)2mg = \omega^2 \cdot \left(m \cdot L + \frac{1}{3} \cdot M \cdot L^2\right)

Finally, isolate ω\omega:

ω2=2mgm⋅L+13⋅M⋅L2\omega^2 = \frac{2mg}{m \cdot L + \frac{1}{3} \cdot M \cdot L^2}

Taking the square root:

ω=2mgm⋅L+13⋅M⋅L2\omega = \sqrt{\frac{2mg}{m \cdot L + \frac{1}{3} \cdot M \cdot L^2}}

Hence, the angular velocity of the rod-object system when the object reaches the end of the rod is:

ω=2mgm⋅L+13⋅M⋅L2\omega = \sqrt{\frac{2mg}{m \cdot L + \frac{1}{3} \cdot M \cdot L^2}}

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