Calculate the velocity of a mass

  • Thread starter Thread starter carr
  • Start date Start date
  • Tags Tags
    Mass Velocity
Click For Summary

Homework Help Overview

The discussion revolves around calculating the velocity of a mass in a system involving two masses, one on an inclined plane and the other hanging vertically. The problem involves understanding the forces acting on both masses and how they relate to acceleration and velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equations of motion for the two masses and attempt to combine them to find acceleration. There are questions about the validity of the equations used and the relationship between distance covered by the masses and the tautness of the string. Some participants suggest using kinematic equations to find velocity but express uncertainty about the time variable.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the equations and their implications. Some guidance has been offered regarding the use of kinematic equations and the assumption of constant acceleration, but there is no explicit consensus on the correct approach or final outcome.

Contextual Notes

Participants note the challenge of not having the time variable, which complicates the calculation of velocity. There is also mention of the constraints imposed by the problem setup, such as the relationship between the displacements of the connected masses.

carr
Messages
2
Reaction score
0
Member warned about not using the homework template
Hi
I have this question to do :

upload_2016-3-12_16-54-39.png


I want to calculate the velocity ... I did like this:

m2 * a = m2 * g - T
m1 * a = T - μ*m1*g*sin α
combine them
m1 * a = -m2*a + m2*g - μ*m1*g*sin α

Now what should I do ?
thanks
 
Physics news on Phys.org
carr said:
m2 * a = m2 * g - T
m1 * a = T - μ*m1*g*sin α
combine them
m1 * a = -m2*a + m2*g - μ*m1*g*sin α

Now what should I do ?
thanks

why you ask? the vertical mass can cover only a finite distance ... so the inclined plane -mass should cover the same distance as string/cable is taut... then from above equations you can calculate the acceleration so you know speed... well i may be wrong... but one can try.
 
drvrm said:
why you ask? the vertical mass can cover only a finite distance ... so the inclined plane -mass should cover the same distance as string/cable is taut... then from above equations you can calculate the acceleration so you know speed... well i may be wrong... but one can try.

ok the acceleration is

a=(m2*g - μ*m1*g*sin α)/(m1+m2)
and v=a/t
but I do not have t (time)
 
carr said:
ok the acceleration is

a=(m2*g - μ*m1*g*sin α)/(m1+m2)
and v=a/t
but I do not have t (time)

are you correct in writing above equation for 'a' and v ? think...
connected systems have same displacements, velocity-magnitude and accelerations. v=a/t is not correct dimensionally - as it is accelerated motion with initial vel=0 one can use v= sqrt(2.acceleration. distance covered)
 
carr said:
ok the acceleration is

a=(m2*g - μ*m1*g*sin α)/(m1+m2)
and v=a/t
but I do not have t (time)

If the acceleration ##a## is constant, you have a simple problem of an object falling with constant acceleration. The acceleration happens to be different from ##g## in this case, but that does not matter; the equations are exactly the same as those you must have seen already in previous courses; if not, you can solve the differential equation for the motion., to get ##x(t)##.
 
carr said:
ok the acceleration is

a=(m2*g - μ*m1*g*sin α)/(m1+m2)
and v=a/t
but I do not have t (time)

In that case I believe that you could use the kinematic equation V2 = V02+2ax
Assuming that the acceleration is constant which it should be since it is caused by gravity in this case
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 19 ·
Replies
19
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
3
Views
3K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K