a) consider an element of length dz and mass dm at a distance of z from mass m on the rod. Force due to this element will be Gmdm/z2. dm = Mdz/L.
Now for total mass integrate it from x = z to x = z+L.
F = GMm/z(z+L). Towards - z direction.
b). Consider an element of length dz at a distance of z from the upper end of lower rod having mass dm.
F on that dm will be GMdm/z(z+L).
DM will be again Mdz/L.
Now integrate it from z= L to z= 2L.
Consider a thin uniform rod of mass M and length L, positioned along the z-axis with...
A uniform thin rod of mass M and length L lies on the positive x-axis with one end at the origin.Consider an element of the rod of length dx., and mass dm at point where 0<x<L. a) What is the gravitational field produced by the mass element of any value of X? b)Calculate the total gravitational field produced by the rod. C)Find the gravitational force on a point particle of mass m0 at x0. D) Show that for x0>>L the...
A thin, uniform rod has length L and the linear density a (i.e. total mass M=al). A point mass m is placed at distance x from one end of the rod, along the axis of the rod. Calculate the gravitational force of the rod on the point mass m. (Hint: element of the mass is dM = adx) -GmM/x? O-GmM/(L2-x2) -GmM/(x+.5L) -GmM/(x2+Lx)
20. A thin rod of mass M and length L gravitationally interacts with a point mass m that is a perpendicular distance a away from its left end (see the figure). The rod is non-uniform, and its linear density (mass per unit length) increases with the distance from its left end according to 2(x) = 2Mx/L?, where x is the horizontal coordinate along the rod (so that x = 0 is at its left end and x=L is at its...
A very thin, straight, uniform rod has a length of 3.00 m and a total mass of 7.00 kg. Treating the rod as essentially a line segment of mass (distributed uniformly), do the following: (i) Use integration to prove that the rod's center of mass is located at its center point. (Reminders: dmnds mass (and that axis is perpendicular to the rod). with the previous result-to calculate lemr the moment of inertia of the rod about an axis through one...
A thin rod of length L lies along the x-axis. It has a uniform linear charge distribution λ0. a) What is the value of the electric potential at a given point x located to the right of the rod? Take V=0 at infinity.b) What is the strength of the electric field at the point x?
A uniform rigid thin rod of length 1 m and mass 4.99 kg is moving on a horizontal xy plane. At a certain instant when the rod is along the y-axis, the instantaneous velocities of the two ends of the rod, A and B, are (4.92 m/s) and (14.40m/s) , respectively. What is the total kinetic energy of the rod at this moment in units of J?
(a) Knowing that the moment of inertia of a thin uniform metallic rod of mass m and length L about an axis through its center of mass is (1/12) ml?, what is its moment of inertial about a parallel axis through one of its ends (show your calculation). (b) A physical pendulum consisting of a thin metallic rod of mass m = 200.0 g and of length L = 1.000 m is suspended from the upper end by a frictionless...
1. A thin rod of length L and total mass M has a linear mass density that varies with position as λ(x)-γ?, where x = 0 is located at the left end of the rod and γ has dimensions M/L3. ĮNote: requires calculus] (a) Find γ in terms of the total mass M and the length L. (b) Calculate the moment of inertia of this rod about an axis through its left end, oriented perpen dicular to the rod; expressed...
(a) Knowing that the moment of inertia of a thin uniform metallic rod of mass m and length L about an axis through its center of mass is (1/12) mL?. what is its moment of inertial about a parallel axis through one of its ends (show your calculation). (b) A physical pendulum consisting of a thin metallic rod of mass m = 200.0 g and of length L - 1.000 m is suspended from the upper end by a frictionless...
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