Linear Density & Center of Mass

AI Thread Summary
To determine the mass of a rod with a linear density function D = 50.0g/m + 20.0xg/m², integration is required. The mass M can be calculated using the integral M = ∫_0^L D(x') dx' from 0 to 0.3 m. The center of mass is found using the formula x_{CM} = (1/M) ∫_0^L x' D(x') dx'. Both integrals need to be evaluated to find the mass and the center of mass accurately. Proper integration will yield the necessary results for both parts of the question.
Ineedphysicshelp
Messages
7
Reaction score
0
Would anyone be able to give me a hint for this question? Do I take the integral? I have possible answers but I am unsure.

A rod of length 30.0 cm has a linear density given by:

D = 50.0g/m + 20.0xg/m^2

where x is the distance from one end measured in meters. a. What is the mass of the rod. b. how far from the x=0 end is the center of mass?
 
Physics news on Phys.org
Yes, you have to do the integration from 0 to 0.3 m.

Linear density (m') is mass/(unit length), so mass M = \int{m'(x)}\,dx.
 
Last edited:
You'll just integrate to find the mass:

M = \int_0^L D(x') dx'

and the center of mass is given by

x_{CM} = \frac {1}{M} \int_0^{L} x' D(x') dx'
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top