Question

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L and mass 2M. Each rod then has one of its supporting strings cut, causing the rod to begin pivoting about the end that is still tied up. Which rod has a larger initial angular acceleration?A Rod 1 B. Rod 2 C. The initial angular acceleration is the same for both.

Answer 1

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is  

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

Answer 2

The correct option is option (A).

The initial angular acceleration of rod 1 is larger than that of the rod 2.

Angular acceleration:

Rod 1 has mass M and length L,

whereas Rod 2 has mass 2M and length 2L.

Considering the rods to be uniform, their center of mass will be at the midpoint of the rod.

So the torque acting on Rod 1:

τ₁ = Mg(L/2) = I₁α₁

where I₁ = (1/3) ML² is the moment of inertia of rod 1

and α₁ is the angular acceleration of the rod 1

Mg(L/2) = (1/3) ML²α₁

α₁ = 3g/2L

Similarly, for Rod 2:

τ₂ = (2M)g(L) = I₂α₂

where I₂ = (1/3) 2M(2L)²

I₂ = (8/3)ML² is the moment of inertia of rod 2

and α₂ is the angular acceleration of the rod 2

2MgL = (8/3)ML²α₂

α₂ = 3g/4L

Now,   3g/2L > 3g/4L

thus, α₁ > α₂

Learn more about angular acceleration:

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