Mechanics ynamics (Newton's 2nd Law?)

Click For Summary
SUMMARY

The discussion focuses on applying Newton's 2nd Law to analyze a system involving a 5 kg sphere and a 1 kg block connected by a rigid rod. The participants confirm that at the moment of release, both the initial velocity and the normal acceleration are zero. The tension in the rod is calculated using the formula T = (m_s)(g)(sin(30)), yielding a value of 24.525 N. Additionally, the horizontal component of tension is derived, leading to T_x = 21.239 N, which is used to find the acceleration of the block.

PREREQUISITES
  • Understanding of Newton's 2nd Law of Motion
  • Knowledge of free body diagrams and force analysis
  • Familiarity with concepts of relative acceleration
  • Basic principles of circular motion and tension in strings
NEXT STEPS
  • Study the derivation of coupled differential equations in dynamics
  • Learn about the application of Work-Energy Theorem in mechanical systems
  • Explore the concept of relative motion in connected systems
  • Investigate the effects of different angles on tension and acceleration in similar setups
USEFUL FOR

Students studying mechanics, physics educators, and anyone interested in solving dynamics problems involving connected objects and forces.

Oblivion77
Messages
113
Reaction score
0
Mechanics:Dynamics (Newton's 2nd Law?)

Homework Statement


ic0ot2.png


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?
 
Physics news on Phys.org


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?
 


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?
 


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.
 


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.
 


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?
 


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.
 


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


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

A_x is the acceleration in the x direction of the sphere.
A_y is the acceleration in the y direction of the sphere.
a_x is the acceleration in the x direction of the block.

Find A_x. A_y, a_x as a function of T, the tension in
the rope.

Find another relation between A_x. A_yand a_x
by differentiating (X(t)-x(t))^2 + Y(t)^2 = L^2.

Solve for T
 

Similar threads

Replies
3
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
6K
Replies
13
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
16
Views
3K
Replies
4
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
7K