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.
For Further Reading
* Physics - KE - bobpursley, Tuesday, March 27, 2007 at 8:21pm
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
*******************
I am thinking that maybe I solved for the velocity incorrectly. Do I use V=sqrt(gR) or do I have to take into account the center of mass and do Vcom and do an additional step under the sqrt???
To solve part (a) of the problem, we can use the principle of conservation of energy.
At the top of the loop, the gravitational potential energy is converted into both kinetic energy and rotational kinetic energy.
The gravitational potential energy at height h is given by mgh, where m is the mass of the ball and g is the acceleration due to gravity.
The kinetic energy can be expressed as (1/2)mv^2, where v is the velocity of the ball at the top of the loop.
The rotational kinetic energy can be expressed as (1/2)Iω^2, where I is the moment of inertia of the ball (given by (2/5)MR^2) and ω is the angular velocity of the ball.
At the top of the loop, the ball is in pure rolling motion, so the linear velocity v is related to the angular velocity ω by v = ωR.
Using these relationships, we can write the conservation of energy equation as:
mgh = (1/2)mv^2 + (1/2)Iω^2
Substituting ω = v/R and I = (2/5)MR^2, we have:
mgh = (1/2)mv^2 + (1/2)((2/5)MR^2)(v/R)^2
Simplifying the equation, we get:
gh = (1/2)v^2 + (1/5)Mv^2
Combining like terms, we have:
gh = (7/10)mv^2
To ensure that the ball does not leave the track at the top of the loop, the centripetal force must be greater than or equal to the gravitational force:
mv^2/R >= mg
Substituting mv^2 = gh, we have:
gh/R >= mg
Simplifying the equation, we get:
h >= R + R/7
Given that R = 2.16 m, we can solve for the minimum required height h:
h >= 2.16 + 2.16/7
h >= 2.5029 m
So, the ball must be released from a minimum height of 2.5029 meters above the bottom of the track to ensure that it does not leave the track at the top of the loop.
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.
At point Q, the ball is still experiencing a gravitational force acting downwards and a normal force acting towards the center of the loop.
The horizontal component of the force can be found using Newton's second law for rotational motion, which states that the sum of the torques acting on an object must equal the moment of inertia times the angular acceleration.
Since the ball is in pure rolling motion, the angular acceleration α is related to the linear acceleration a by α = a/R, where R is the radius of the loop.
The sum of the torques acting on the ball is given by τ = Iα, where τ is the torque and I is the moment of inertia.
At point Q, the torque due to the net horizontal force is equal to the torque due to the friction force, since there is no other torque acting on the ball.
Therefore, we have:
FhR = μmgR
Simplifying the equation, we get:
Fh = μmg
where μ is the coefficient of friction.
Therefore, the magnitude of the horizontal component of the force at point Q is μmg.