Solving Part B: Acceleration of End B after String Cut

[CaptFormal]

Homework Statement

A rod of length 69.0 cm and mass 1.70 kg is suspended by two strings which are 43.0 cm long, one at each end of the rod.

http://schubert.tmcc.edu/res/msu/physicslib/msuphysicslib/20_Rot2_E_Trq_Accel/graphics/prob13a_1016full.gif

(A) The string on side B is cut. Find the magnitude of the initial acceleration of end B.

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The string on side B is retied and now has only half the length of the string on side A.

http://schubert.tmcc.edu/res/msu/physicslib/msuphysicslib/20_Rot2_E_Trq_Accel/graphics/prob13b_1016half.gif

(B) Find the magnitude of the initial acceleration of the end B when the string is cut.

Homework Equations

The Attempt at a Solution

A bit lost on even how to solve this one. Any assistance would be appreciated. Thanks.

 

[Monocerotis]

Assuming no air resistance
Fg = (m)(g)
Fg = (1.70kg)(9.8m/s^2)
Fg = 16.6

a = Fnet/m
a = 16.6/1.70
a = 9.8 m/s^2

Think that's right...?

 

[tomcue]

I would approach this problem by first identifying the relevant equations and principles involved. In this case, we can use the equations for rotational motion and Newton's laws of motion.

For part A, we can use the principle of conservation of angular momentum to find the initial acceleration of end B after the string is cut. This principle states that the initial angular momentum of a system will be equal to the final angular momentum of the system, as long as there are no external torques acting on the system.

In this case, the initial angular momentum of the system is zero, since the rod is initially at rest. After the string is cut, the rod will start to rotate around its center of mass, with end B moving downwards. The final angular momentum of the system will be the angular momentum of the rod rotating around its center of mass, which can be calculated using the equation L = Iω, where I is the moment of inertia and ω is the angular velocity.

We can then equate the initial and final angular momenta to find the initial angular velocity of the rod, which can then be used to calculate the initial acceleration of end B.

For part B, we need to take into account the fact that the string on side B is now half the length of the string on side A. This means that the moment of inertia of the system will also change, as the distribution of mass along the rod will be different. We can use the equation for the moment of inertia of a rod rotating around its center of mass (I = (1/12)ML^2) to calculate the new moment of inertia.

Once we have the new moment of inertia, we can repeat the same steps as in part A to find the initial acceleration of end B.

Overall, the key to solving this problem is to carefully consider the principles and equations involved, and to apply them correctly to the situation at hand.