Mechanics ynamics (Newton's 2nd Law?)

In summary, the problem involves a sphere S with a mass of 5 kg attached by a rigid rod to a 1 kg block B, which is free to slide with no friction in a horizontal slot. The system is released from rest and the question asks for the tension in the rod and the accelerations of both blocks at the instant of release. To solve this problem, one can use Newton's 2nd law and conservation of energy. The initial velocity can be assumed to be zero, and work-energy conservation can be used to find the acceleration after a certain time interval. However, relative acceleration needs to be considered in this problem. By relating the forces and finding a pair of coupled differential equations with tension as a variable, one can
  • #1
Oblivion77
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Mechanics:Dynamics (Newton's 2nd Law?)

Homework Statement


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A sphere S with a mass of 5 kg is attached by a rigid rod to 1 kg block B which is free to slide with no friction in a horizontal slot. The system is released from rest. At the instant when it is released, find the tension in the rod and the accelerations of both blocks.

Homework Equations



Sum of the forces in x = ma
Sum of the forces in y = ma

The Attempt at a Solution



I am not exactly sure how to tackle this problem. Would I need to solve this using Newton's 2nd law? Or would this question require conservation of energy?
 
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  • #2


Using Newton's 2nd Law would definitely help.
And yes, your equations are correct.

Can u show your approach and the answer you are getting?
 
  • #3


I have another question then. Since the problem is asking "the instant it is released from rest" can you assume the velocity is 0 at that instant? Therefore the component of normal acceleration is 0?
 
  • #4


Oblivion77 said:
I have another question then. Since the problem is asking "the instant it is released from rest" can you assume the velocity is 0 at that instant? Therefore the component of normal acceleration is 0?

Yes, you can say that the initial velocity would be zero.. and yes, the normal acc.(due to circular motion) is also zero.
I guess you also have to use Work-Energy Conservation to solve.
 
  • #5


blitz.km said:
Yes, you can say that the initial velocity would be zero.. and yes, the normal acc.(due to circular motion) is also zero.
I guess you also have to use Work-Energy Conservation to solve.

I don't think work and energy can be used here since you are looking for the accelerations.

Would relative acceleration need to be considered for this problem since the acceleration of the sphere is moving while the block is moving.
 
  • #6


Oblivion77 said:
I don't think work and energy can be used here since you are looking for the accelerations.
Umm.. yeah.
But we would have used work-energy in case we needed acceleration after some time interval.

Oblivion77 said:
Would relative acceleration need to be considered for this problem since the acceleration of the sphere is moving while the block is moving.
YES.

Can u post your complete approach?
 
  • #7


You need to relate the forces that act between the block and the sphere. I would pick the side of the wall as your reference frame, x.
I think you should get a pair of coupled differential equations with tension in there somewhere. Then use your initial conditions.
 
  • #8


Would tangential acceleration of S = (m_s)*g*cos(30)=(m_s)(a_st)
so a_s tangential = 8.50 m/s^2?
 
  • #9


and also since system is initially at rest wouldn't T=(m_s)*(g)*(sin(30)) = 24.525N
then solving for T in the horizontal dir. T_x=T(cos(30))

so T_x=(m_s)*(g)*(sin(30))*(cos(30)) =21.239
(T_x)/(m_b)= (a_b) = 21.239 m/s^2
 
  • #10


clampala said:
and also since system is initially at rest wouldn't T=(m_s)*(g)*(sin(30)) = 24.525N
then solving for T in the horizontal dir. T_x=T(cos(30))

so T_x=(m_s)*(g)*(sin(30))*(cos(30)) =21.239
(T_x)/(m_b)= (a_b) = 21.239 m/s^2

You need to take relative acceleration into account I am pretty sure.
 
  • #11


Let X, Y be the position of the sphere, and x the x-position of the block

[itex] A_x [/itex] is the acceleration in the x direction of the sphere.
[itex] A_y [/itex] is the acceleration in the y direction of the sphere.
[itex] a_x [/itex] is the acceleration in the x direction of the block.

Find [itex] A_x [/itex]. [itex] A_y [/itex], [itex] a_x [/itex] as a function of T, the tension in
the rope.

Find another relation between [itex] A_x [/itex]. [itex] A_y [/itex]and [itex] a_x [/itex]
by differentiating (X(t)-x(t))^2 + Y(t)^2 = L^2.

Solve for T
 

What is Newton's 2nd Law of Motion?

The 2nd Law of Motion, also known as the Law of Acceleration, states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In simpler terms, it means that the greater the force applied to an object, the greater its acceleration will be, and the more massive an object is, the less it will accelerate.

How is Newton's 2nd Law mathematically expressed?

The mathematical expression of Newton's 2nd Law is F = ma, where F represents the net force, m represents the mass of the object, and a represents the acceleration. This formula can be rearranged to solve for any of the three variables.

What are some real-life examples of Newton's 2nd Law in action?

A few examples of Newton's 2nd Law in everyday life include pushing a shopping cart, throwing a ball, and riding a bicycle. In all of these scenarios, the force applied to the object causes it to accelerate in the direction of the force.

How does Newton's 2nd Law relate to inertia?

Inertia is the tendency of an object to resist changes in its state of motion. Newton's 2nd Law is directly related to inertia, as the greater the mass of an object, the more inertia it has, and therefore, the more force is required to accelerate it.

What is the difference between mass and weight in the context of Newton's 2nd Law?

Mass and weight are often used interchangeably in everyday language, but they have different meanings in the context of Newton's 2nd Law. Mass is a measure of an object's inertia, while weight is a measure of the force of gravity acting on an object. In other words, an object's mass does not change, but its weight can vary depending on the strength of the gravitational field it is in.

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