Three objects of uniform density—a solid sphere, a solid cylinder, and a hollow cylinder— are placed at the top of an incline (f
ig. cq10.13). they are all released from rest at the same elevation and roll without slipping. (a) which object reaches the bottom first
1 answer:
First you must make a free body diagram of each body to find the acceleration of each one. After you find the general formula of acceleration, you must replace the values of K for each body. (Assuming they all have the same mass). Finally, by replacing K, the body that gets the most acceleration will be the first to reach the bottom.I attached the solution.In this case the solid sphere reaches the bottom first.
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So the correct ans is B.
hope it helps u.
The concept of conservation of energy and harmonic motion allows to find the result for the power where the kinetic and potential energy are equal is:
x = 0.135 cm
Given parameters
- The mass m = 220 g = 0.220 kg
- The spring cosntnate3 k = 7.0 N / m
- Initial displacement A = 5.2 cm = 5.2 10-2 m
To find
- The position where the kinetic and potential energy are equal
A simple harmonic movement is a movement where the restoring force is proportional to the displacement, the result of this movement is described by the expression.
x = A cos wt + fi
w² =
Where x is the displacement from the equilibrium position, A the initial amplitude of the system, w the angular velocity t the time, fi a phase constant determined by the initial conditions, k the spring constant and m the mass.
The speed is defined by the variation of the position with respect to time.
v =
let's evaluate
v = - A w sin (wt + Ф)
Since the body releases for a time t = 0 the velocity is zero, therefore the expression remains.
0 = - A w sin Ф
For the equality to be correct, the sine function must be zero, this implies that the phase constant is zero
x = A cos wt
Let's find the point where the kinetic and potential energy are equal.
K = U
½ m v² = m g x
we substitute
½ A² w² sin² wt = g A cos wt
sin² wt = cos wt
let's calculate
w =
w = 5.64 rad / s
sin² 5.64t = 2 9.8 / 0.052 cos 5.64t
sin² 5.64t = 376.92 cos 5.64 t
1 - cos² 5.64t = 376.92 cos 5.64t
cos² 5.64t -376.92 cos564t -1 = 0
we make the change of variable
x = cos 5.64t
x²- 376.92 x - 1 = 0
x = 0.026
cos 5.64t = 0.026
Let's find the displacement for this time
x = 5.2 10-2 0.026
x = 1.35 10-3 m
In conclusion Using the concepts of conservation of energy and harmonic motion we can find the result for the could where the kientic and potential enegies are equal is:
x = 0.135 cm
Learn more here: brainly.com/question/15707891