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zoom1
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Didn't get this equation which is written for the pendulum part of the cart.
Must be easy but couldn't get it.
Understanding the Pendulum Dynamics Equation for a Cart
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The discussion focuses on understanding the pendulum dynamics equation for a cart, specifically how it relates to the forces acting on the center of mass of the rod. The equation involves terms for tangential acceleration (##\ddot\theta##) and radial (centripetal) acceleration (##\dot\theta^2##), which are derived from basic principles of force and acceleration. There is clarification that the length of the rod should be considered as 2l instead of l in the equation. Additionally, the centripetal force is noted to act radially, but the equation specifically addresses the force component in the x direction. Overall, the discussion aims to clarify the relationship between these forces and their representation in the equation.
Didn't get this equation which is written for the pendulum part of the cart.
Must be easy but couldn't get it.
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AlephZero
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It is just "force = mass x acceleration" in the x direction, for the center of mass of the rod. (The length of rod isn't shown on your diagram, but if the equation is correct it is 2l, not l).
The ##\ddot\theta## term is from the tangential acceleration of the rod.
The ##\dot\theta^2## term is the from the radial (centripetal) accleration.
The ##\ddot\theta## term is from the tangential acceleration of the rod.
The ##\dot\theta^2## term is the from the radial (centripetal) accleration.
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zoom1
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AlephZero said:
It's a little clearer now, but didn't understand the centripetal force component.
Isn't it
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AlephZero
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Yes.
You also know ##v = \omega r## and ##\omega = \dot\theta##
The centripetal force is radial (along the length of the rod), but the equation is for the component of the force in the X direction.
You also know ##v = \omega r## and ##\omega = \dot\theta##
The centripetal force is radial (along the length of the rod), but the equation is for the component of the force in the X direction.
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19.
For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Let's declare that for the cylinder,
mass = M = 10 kg
Radius = R = 4 m
For the wall and the floor,
Friction coeff = ##\mu## = 0.5
For the hanging mass,
mass = m = 11 kg
First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on.
Force on the hanging mass
$$mg - T = ma$$
Force(Cylinder) on y
$$N_f + f_w - Mg = 0$$
Force(Cylinder) on x
$$T + f_f - N_w = Ma$$
There's also...
This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is
The second part of the problem is exactly why you get this affect.
And one more spoiler:
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