In Figure 11-32 (which shows a ball at the top of an incline, at the bottom of the incline a loop begins with radius R and Q a point on the loop lined up with the center of the loop), a solid brass ball of mass m and radius r will roll without slipping along the loop-the-loop track when released from rest along the straight section.
(a) From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop? (The radius of the loop-the-loop is R = 2.16 m. Assume R>>r, and the mass of the ball is 3.2 kg.)
(b) If the brass ball is released from height 6R above the bottom of the track, what is the magnitude of the horizontal component of the force acting on it at point Q?
*My work so far*
for finding I for a sphere I believe that I need to use I = (2/5)MR^2 which is I = 5.97 kg*m^2. Can I use V = sqrt(gR) to find the V...if so, V = 4.6 m/s. I am not sure what to do now. Can I use -Fs = m(-V^2/R)???
For Further Reading
* Physics - KE/rotation - drwls, Monday, March 26, 2007 at 5:06am
To stay on track at the top of the loop, MV^2/R must equal or exceed Mg there. Use conservation of enewrgy to relate the velocity there to the distance of the release point above the bottom of the loop.
The total kinetic energy when the velocity is V is
KE = (1/2) M V^2 + (1/2)(2/5)Mr^2*(V/r)^2 = (3/5) M V^2
If H is the initial height of the ball above the bottom of the loop, then at the top of the loop, 2R from the bottom,
MgH = (3/5) M V^2 + 2 MgR
V^2 = (5/3) gH -(10/3)gR > g
Solve for the minimum required H
My work:
V^2 = (5/3)*gH - (10/3)gR...I am not sure what he meant by > g in the equation above???
using V^2 = (5/3)*gH - (10/3)gR I found H to be 5.6155 meters which is the wrong answer. Any thoughts??? Thanks.
Mv^2/R >= Mg to stay on the top of the loop.
But mv^2 = 5/3 MgH -10/3 gR from
MgH = (3/5) M V^2 + 2 MgR
Therefore,
5/3 MgH -10/3 MgR >= MgR
divide through by Mg
5/3 H >= R+10R/3
or H >= 5.616M or to three places 5.62m
To solve part (a) of the problem, we need to determine the minimum height (H) from which the marble should be released to ensure it does not leave the track at the top of the loop.
We start by considering the conditions required for the marble to stay on the track at the top of the loop. At the top, the centripetal force provided by the normal force must be equal to or greater than the gravitational force pulling the marble downward. Mathematically, this can be expressed as:
M * V^2 / R >= M * g
Where M is the mass of the ball, V is the velocity of the ball at the top of the loop, R is the radius of the loop (given as 2.16m), and g is the acceleration due to gravity.
Next, we need to find an equation that relates the velocity (V) to the height (H) from which the ball is released. We can do this by using the principle of conservation of energy. At the topmost point of the loop, the total kinetic energy of the ball is equal to the potential energy it had when released from the height H.
The total kinetic energy can be calculated by considering the translational kinetic energy and the rotational kinetic energy. Since the ball is rolling without slipping, the rotational kinetic energy is given by (1/2) * (2/5) * M * r^2 * (V/r)^2, where r is the radius of the ball. The translational kinetic energy is given by (1/2) * M * V^2.
Thus, the total kinetic energy is given by:
KE = (1/2) * M * V^2 + (1/2) * (2/5) * M * r^2 * (V/r)^2
= (3/5) * M * V^2
Setting the total kinetic energy equal to the potential energy at height H, we have:
(3/5) * M * V^2 = M * g * H
Simplifying, we obtain:
V^2 = (5/3) * g * H
Now, we substitute this expression for V^2 in the inequality M * V^2 / R >= M * g:
(5/3) * g * H / R >= g
Canceling out the g, we find:
(5/3) * H / R >= 1
Rearranging the inequality, we have:
H >= R * 3/5
Now substituting the given value R = 2.16m, we find:
H >= (2.16m) * (3/5) = 1.296m
Therefore, the minimum height from which the marble should be released is 1.296 meters.
For part (b) of the problem, we need to find the magnitude of the horizontal component of the force acting on the ball at point Q when it is released from a height of 6R above the bottom of the track.
To solve this part, we need to consider the forces acting on the ball at point Q. At this point, the only horizontal force is the normal force, which provides the required centripetal force to keep the ball moving in a circular path.
The magnitude of the centripetal force is given by:
F = M * V^2 / R
Where M is the mass of the ball, V is the velocity of the ball at point Q, and R is the radius of the loop.
Using the equation we derived earlier for the velocity at the top of the loop, V^2 = (5/3) * g * H, and substituting the given values M = 3.2kg, R = 2.16m, and H = 6R, we can calculate the magnitude of the centripetal force:
F = (3.2kg) * [(5/3) * (9.8m/s^2) * (6 * 2.16m)]^2 / (2.16m)
= (3.2kg) * (5 * 9.8m/s^2 * 12.96m)^2 / (2.16m)
= (3.2kg) * (9.8m/s^2 * 155.52m)^2 / (2.16m)
= (3.2kg) * (9.8m/s^2 * 155.52m)^2 / (2.16m)
Calculating this expression, we find the magnitude of the horizontal component of the force acting on the ball at point Q. Note that the value will depend on the given values for H and R.