Physics Olympiad Problem #001 - Period of Solid Cylinder-Spring System
A solid cylinder with mass m = 500 g is placed on a board with mass M = 100 g connected to a spring with constant k = 20 N/m. The cylinder rolls without slipping when moving relative to the board. If the board is given a small deviation, calculate the period of oscillation of the board?
#IPHO #PhysicsOlympiad
#IPHO #PhysicsOlympiad
Category
📚
LearningTranscript
00:00Hi friends, good afternoon.
00:04Afternoon is indeed for lying down.
00:08This atmosphere is suitable for watching quite complicated physics videos.
00:14You can read this question for a while.
00:24Although there is a short illustration,
00:27there is no harm in watching this short animation.
00:32The following system consists of a board connected to a spring attached to the wall.
00:38It looks like the contact surface between the board and the floor is smooth.
00:43However, the top surface of the board is rough.
00:48There is a solid cylinder there.
00:54This is roughly the three-dimensional view of this system.
01:01Now, the board is slightly deviated to the left, then released.
01:05The board will perform simple harmonic motion.
01:11What is interesting about this question is that the cylinder rolls without slipping when moving relative to the board.
01:19Now, we are asked to calculate the oscillation period of the board.
01:26Visually, this is a complicated question.
01:31Whatever it is, let's discuss it.
01:35As with other simple harmonic motion questions, we have to draw a diagram of the forces acting on the system.
01:43This system consists of three different objects.
01:47To make it easier to analyze, we will separate this system into two objects with mass.
01:54A board with index 1 and a solid cylinder with index 2.
02:01We start from the solid cylinder.
02:03The solid cylinder is around the Earth.
02:06There is the Earth's gravitational force.
02:11The cylinder is in contact with the board.
02:14There is also a normal force and two upwards.
02:19During rolling, the surface of the cylinder will rub against the surface of the board.
02:24The friction force has a direction opposite to the direction of translational motion.
02:29It seems that this is the only force acting on the solid cylinder.
02:35For the board, there is no single force.
02:39The force appears as an action-reaction pair.
02:42On the board will work the friction force Fs and N2 in the opposite direction.
02:49As an interaction with the Earth, the board also works the gravitational force N1g.
02:57In interaction with the floor surface, there is a normal force N1.
03:04The stretched spring will exert a spring force Kx1.
03:11Before analyzing these forces, we will review the special requirements in this problem.
03:18The cylinder rolls without slipping.
03:20We can draw some points on the surface of the cylinder and on the board.
03:27The cylinder will not slip when each point is in line with each other.
03:32This is correct.
03:35How to formulate this mathematically?
03:40Suppose the board is accelerating to the right with an acceleration of A1 to the right.
03:48The cylinder is rotating with an angular acceleration of α2.
03:54Since the cylinder is rolling, there are two accelerations at the point of contact.
03:59A translational acceleration of A2 to the left and a tangential acceleration of α2 are to the right.
04:07A1 and A2 are the accelerations measured by an observer on the floor surface.
04:15In order for each point displacement on the two surfaces to correspond to each other,
04:20A1 must be equal to minus A2 plus α2r.
04:26This is equation 1.
04:30All of these forces will interact to form a resultant force.
04:36The resultant horizontal force on the board, minus Fs minus kx1 is equal to M1A1.
04:43This is equation 2.
04:47The resultant horizontal force on the cylinder, Fs is equal to minus M2A2,
04:53because the direction of A2 is to the left.
04:58The resultant torque on the rotation of the cylinder, Fsr is equal to half M2r squared α2,
05:04because the moment of inertia of a solid cylinder is half Mr squared.
05:10Let's call these two as equation 3 and equation 4.
05:18We have four equations, with four unknown quantities.
05:22A1, A2, Fs, and α2.
05:27It seems that these mathematical equations are enough.
05:33The next step only requires mathematical skills.
05:38We can write equation 1 as α2r equals A1 plus A2.
05:46The equation that contains the value of α is equation 4.
05:50We can write equation 4 like this.
05:56We can cross out the quantity r on both sides.
06:00From here, Fs is equal to half M2 multiplied by A1 plus A2.
06:08In equation 3, the magnitude of Fs is seen.
06:12What if we substitute the value of Fs from equation 4 into equation 3?
06:20We can cross out M2.
06:24Next, arrange the physical quantities based on similar physical quantities.
06:31From here, A2 is equal to minus one-third A1.
06:37Knowing the value of A2, we can calculate the friction force as a function of A1.
06:42Why? Because what is requested is the period of oscillation of the board.
06:47The acceleration that is identical to the board is A1.
06:52So we will substitute the acceleration of A2 into the friction force equation.
07:00From here, Fs is equal to one-third M2 A1.
07:06Now, we can use this value of Fs into equation 2.
07:13We can write equation 2 like this.
07:19Are you familiar with this equation?
07:22That's right. This equation is identical to the general equation of simple harmonic motion.
07:30So, we have to change the equation to fit the general equation.
07:38Omega is nothing but the square root of 3k per 3 M1 plus M2.
07:45If you still remember, omega is 2 pi over big T.
07:50This is the equation of the oscillation period from the board.
07:56It seems that all the values have been written on the question sheet.
08:02Writing in one decimal place, T is around 0.7 seconds.
08:08How complicated is this question?
08:12Happy learning everyone!