A small marble of mass m = 0.01 kg. and radius r = 0.5 cm is attached to a sprin

Question

A small marble of mass m = 0.01 kg. and radius r = 0.5 cm is attached to a spring of spring constant k = 5 N/m. You compress the spring by x = 18 cm with the marble, and then let go. a) If the marble stays attached to the spring, write down the equation for the position of the marble as a function of time, if you let go at time t = 0. Now, say that the marble is launched from the spring instead. It rolls without slipping along the flat ground, until it reaches a mound. The mound can be modeled as a straight ramp up to a part of a circle of radius R_i, like shown in the diagram. The marble doesn't lose any energy when it hits the edge of the mound. The moment of inertia for the marble is I = 2/smr^2. b) What is the minimum radius R_i, such that the ball stays in contact with the mound at all times? c) Does your answer increase or decrease if it's a box that slides without friction, rather than a rolling marble?

Explanation / Answer

part a:

angular frequency=w=sqrt(k/m)=22.36 rad/s
equation of motion will be given by

x(t)=0.18*cos(22.36*t+phi)

where phi is phase constant

at t=0, x=-18 cm=-0.18 m

==>phi=pi rad

hence equation of motion is given by

x(t)=0.18*cos(22.36*t+pi)

part b:

in order to stay in contact with the mound all the time,

the total downward force on the marbel at the highest position should be greater than or equal to 0

for minimum value of R1, the force on the marbel at the highest location is zero.

==>weight-centripetal force=0

==>centripetal force=weight

let speed at top of the circle is v m/s.

then mass*v^2/R1=m*g

==>R1=v^2/g....(1)

in order to find value of v,

we will use conservation of energy principle.

as there is no friction involved,
total energy at the beginning=total energy at the top of the mound

==>0.5*k*x^2=0.5*m*v^2+0.5*I*(v/R1)^2

==>0.081=0.005*v^2+10^(-7)*(v^2/R1^2)

as R1=v^2/g

==>0.081=0.005*v^2+10^(-7)*(v^2/(v^4/g^2))

==>0.081=0.005*v^2+(9.604*10^(-6)/v^2)

==>0.005*v^4-0.081*v^2+9.604*10^(-6)=0

solving for v^2, we get v^2=16.2

==>v=4.025 m/s

==>R1=v^2/g=1.653 m

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