Find the Minimum Height Needed to Complete a Loop-the-Loop Track

[Punchlinegirl]

A solid marble of mass m=35 and radius r=7 cm will roll without slipping along the loop-the-loop track if it is released from rest somewhere on the straight section of the track. From what minimum height h above the bottom of the track must the marble be released to ensure that it doesn't leave the track at the top of the loop? The radius of the loop-the loop is R=1.30 m.

I don't know where to begin. Any advice?

 

[UrbanXrisis]

there needs to be enough centripetal force on the top of the loop to even out the weight so it the mass doesn't fall

\frac{mv^2}{r}=mg

 

[Punchlinegirl]

I don't know the velocity or acceleration

 

[UrbanXrisis]

you find that by

mgh=.5mv^2

find v, plug that into the first equation and solve for h

 

[Skomatth]

Urban, your 2nd equation will not work because you have neglected the rotational kinetic energy of the marble.

Punchline, use Urban's first equation to determine the minimum velocity that the marble needs at the top of the loop. Then apply conservation of energy to find minimum h. You will have to use
\Delta PE = .5mv^2 + .5I\omega^2

 

[UrbanXrisis]

I don't think they have learned rotational energy yet, so I did not include it, but yes you are correct

 

[xanthym]

Urban, your 2nd equation will not work because you have neglected the rotational kinetic energy of the marble.

Punchline, use Urban's first equation to determine the minimum velocity that the marble needs at the top of the loop. Then apply conservation of energy to find minimum h. You will have to use
\Delta PE = .5mv^2 + .5I\omega^2

(Note: From Skomatth's comments, above equation applies at all points along marble's path referenced to the initial Potential Energy. See summary of solution hints below.)

SOLUTION HINTS (including summary of previous hints):
{TOTAL Energy Required by Marble at Loop's Top} =
= {Linear Kinetic Energy @ Top} + {Rotational Kinetic Energy @ Top} + {Potential Energy @ Top} =
= (1/2)*m*v² + (1/2)*I*ω² + m*g*(2*R) =
= (1/2)*m*v² + (1/2)*{(2/5)*m*r²}*(v/r)² + 2*m*g*R =
= (1/2)*m*v² + (2/10)*m*v² + 2*m*g*R =
= m*{(7/10)*v² + 2*g*R}

The above TOTAL Energy must be obtained from the Potential Energy at an initial height "H" above Loop's bottom given by:
{Initial Required Potential Energy} = m*g*H* =
= m*{(7/10)*v² + 2*g*R}
::: ⇒ H = (7/10)*v²/g + (2*R) ::: (Eq #1)

The velocity "v" is then determined from requiring centripetal force at loop's top to equal the marble's weight "m*g":
m*v²/R = m*g
::: ⇒ v = sqrt{g*R} ::: (Eq #2)

Place Eq #2 into Eq #1 and determine "H" to complete solution.


~~

 

[Skomatth]

All I wrote was that the change in potential energy was equal to the kinetic energy at some point in the marble's path. Indeed there is no term for the marble's potential energy at the top of the loop because the LHS side of the equation accounts for it. Were I to write the LHS in terms of m,g,h and R it would be identical to your equation, though I did not because I'm hesitant to work almost half the problem for punchline.