An object of mass m is released from rest at a height h above the surface of a table. The object slides along the inside of the loop-the-loop track consisting of a ramp and a circular loop of radius R shown in the figure. Assume that the track is frictionless.
The answer for this is 5/2*R
I'm searching for some other solutions..
maybe we could use this as an exchange thread..
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We release an oil drop of radius r in air. The density of the oil is 790 kg/m3. C1 and C2 for 1 atmosphere air at 20∘ C are 3.80 × 10−4 (kg/m)/sec and 0.71 kg/m3, respectively.
How small should the oil drop be so that the drag force is dominated by the linear term in the speed (in lectures we called this Regime I). In this regime, the terminal velocity is mg/C1r. [m is the mass of the drop].
r<<
found it by myself..
vterm << vcrit
m*g/(C1*r) << C1/(C2*r)
m=4/3*pi*r^2*d
d:density of oil
solve for r
To find the smallest size of the oil drop such that the drag force is dominated by the linear term in speed (Regime I), we can start by understanding the expression for the drag force in terms of the oil drop's radius and velocity.
In fluid dynamics, the drag force experienced by an object moving through a fluid is given by the equation:
F_drag = 6πrηv
Where:
F_drag is the drag force,
r is the radius of the object,
η is the dynamic viscosity of the fluid, and
v is the velocity of the object.
For a sphere of radius r moving at high velocities, the drag force is dominated by the linear term in the speed when the viscous force (6πηrv) is much smaller than the gravitational force (mg).
In this case, the gravitational force is given by:
F_gravity = m * g
Where:
m is the mass of the oil drop, and
g is the acceleration due to gravity.
To find the smallest size of the oil drop where the drag force is dominated by the linear term in speed, we need to set the viscous force equal to the gravitational force:
6πηrv = m * g
Rearranging the equation:
v = (m * g) / (6πηr)
Now, in Regime I, the terminal velocity is given by:
v_terminal = m * g / (C1 * r)
To find the smallest size of the oil drop, we want the velocity v to be equal to the terminal velocity v_terminal. Therefore, we can set the two equations equal to each other and solve for the radius r:
(m * g) / (6πηr) = m * g / (C1 * r)
Simplifying the equation:
C1 = (6πη) / (m * g)
Rearranging the equation for the radius r:
r = (6πη) / (C1 * m * g)
Now, to find the smallest size of the oil drop, we want r to be as small as possible, which means taking the limit as r approaches zero (r << 1). This implies that the right-hand side of the equation should also approach zero.
Therefore, the smallest size of the oil drop in which the drag force is dominated by the linear term in speed (Regime I) is when:
r << (6πη) / (C1 * m * g)
In this case, r is much smaller than the expression on the right-hand side of the equation.