The rod oscillates with a period of 0. We are asked to find the torsion constant of the string. We first need to find the moment of inertia. Like the force constant of the system of a block and a spring, the larger the torsion constant, the shorter the period. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant?
Explain your answer. A pendulum clock works by measuring the period of a pendulum. In the springtime the clock runs with perfect time, but in the summer and winter the length of the pendulum changes.
When most materials are heated, they expand. Does the clock run too fast or too slow in the summer? What about the winter? If the period should be one second, but period is longer than one second in the summer, it will oscillate fewer than 60 times a minute and clock will run slow. In the winter it will run fast. With the use of a phase shift, the position of an object may be modeled as a cosine or sine function. If given the option, which function would you choose? Assuming that the phase shift is zero, what are the initial conditions of function; that is, the initial position, velocity, and acceleration, when using a sine function?
How about when a cosine function is used? Some people think a pendulum with a period of 1. True or not, what is the length of such a pendulum? How long does it take a child on a swing to complete one swing if her center of gravity is 4. Two parakeets sit on a swing with their combined CMs At what frequency do they swing? What is its new period? A pendulum with a period of 2. What is the acceleration due to gravity at its new location? Privacy Policy. Skip to main content.
Thus we can eliminate final kinetic energy to get: Plugging in expressions for each variable, we get: Eliminating mass and rearranging for maximum velocity, we get: Where the maximum height is two times the length of the pendulum:. Example Question 17 : Pendulums. Explanation : Since the maximum angle achieved by the pendulum is very small, we can use the follow expression to determine the angle of the pendulum at any time : Note how we are using the cosine function since the pendulum began at its highest point.
Example Question 18 : Pendulums. Explanation : The block experiences its maximum velocity when it is at its lowest point of rotation and its minimum velocity at its highest point of rotation. Therefore, we can use the expression for conservation of energy to solve this problem: If we assume that the initial condition is when the block is at the low point of rotation and assume that that point has a height of 0, then we can eliminate initial potential energy: Now substituting in expressions for each of these: Eliminating mass and multiplying both sides of the expression by 2, we get: Then rearranging for final velocity: Where the height is simply twice the length of the pendulum: Plugging in our values:.
Example Question 19 : Pendulums. Explanation : Since we are given the length of the pendulum and told that it begins at rest and in the horizontal position, we can calculate the maximum velocity of the block as it travels through its lowest point using the expression for conservation of energy: We can eliminate initial kinetic energy since the pendulum begins at rest. Substituting in expressions for each of these: Where the initial height is just the length of the pendulum: Rearranging for final velocity, we get: We can then calculate centripetal acceleration from this: Where the radius is the length of the pendulum:.
Example Question 20 : Pendulums. Explanation : We can begin with the expression for conservation of energy to solve this problem: The block will achieve its maximum velocity at the lowest point of rotation. If we say that this point has a height of 0, we can eliminate initial potential: Plugging in expressions for each of these: Multiplying both sides of the expression by , we get: Now let's say that the final condition is when the block has a velocity of.
Rearranging for the final height, we get: Plugging in our values: This is the height that the block is above our reference point when it reaches the desired speed. First, let's begin with a function that tells us how far below the block is from the horizontal: If this does not make sense, draw it out. Moving on, we can take this expression to develop one that tells us the height of the block: Rearranging for angle: Now plugging in our values:. Copyright Notice.
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Zip Code. Track your scores, create tests, and take your learning to the next level! Top Subjects. Our Company. Varsity Tutors. So in the case of a pendulum, it is the gravity force which gets resolved since the tension force is already directed perpendicular to the motion.
The diagram at the right shows the pendulum bob at a position to the right of its equilibrium position and midway to the point of maximum displacement. A coordinate axis system is sketched on the diagram and the force of gravity is resolved into two components that lie along these axes. One of the components is directed tangent to the circular arc along which the pendulum bob moves; this component is labeled Fgrav-tangent. The other component is directed perpendicular to the arc; it is labeled Fgrav-perp.
You will notice that the perpendicular component of gravity is in the opposite direction of the tension force. You might also notice that the tension force is slightly larger than this component of gravity.
The fact that the tension force Ftens is greater than the perpendicular component of gravity Fgrav-perp means there will be a net force which is perpendicular to the arc of the bob's motion. This must be the case since we expect that objects that move along circular paths will experience an inward or centripetal force. The tangential component of gravity Fgrav-tangent is unbalanced by any other force.
So there is a net force directed along the other coordinate axes. It is this tangential component of gravity which acts as the restoring force. As the pendulum bob moves to the right of the equilibrium position, this force component is directed opposite its motion back towards the equilibrium position. The above analysis applies for a single location along the pendulum's arc. At the other locations along the arc, the strength of the tension force will vary. Yet the process of resolving gravity into two components along axes that are perpendicular and tangent to the arc remains the same.
The diagram below shows the results of the force analysis for several other positions. There are a couple comments to be made. First, observe the diagram for when the bob is displaced to its maximum displacement to the right of the equilibrium position. The tension force Ftens and the perpendicular component of gravity Fgrav-perp balance each other. At this instant in time, there is no net force directed along the axis that is perpendicular to the motion.
Since the motion of the object is momentarily paused , there is no need for a centripetal force. Second, observe the diagram for when the bob is at the equilibrium position the string is completely vertical. When at this position, there is no component of force along the tangent direction. When moving through the equilibrium position, the restoring force is momentarily absent. Having been restored to the equilibrium position, there is no restoring force.
The restoring force is only needed when the pendulum bob has been displaced away from the equilibrium position. You might also notice that the tension force Ftens is greater than the perpendicular component of gravity Fgrav-perp when the bob moves through this equilibrium position. Since the bob is in motion along a circular arc, there must be a net centripetal force at this position. In the previous part of this lesson , we investigated the sinusoidal nature of the motion of a mass on a spring.
We will conduct a similar investigation here for the motion of a pendulum bob. Let's suppose that we could measure the amount that the pendulum bob is displaced to the left or to the right of its equilibrium or rest position over the course of time.
A displacement to the right of the equilibrium position would be regarded as a positive displacement; and a displacement to the left would be regarded as a negative displacement. Using this reference frame, the equilibrium position would be regarded as the zero position. And suppose that we constructed a plot showing the variation in position with respect to time.
The resulting position vs. Similar to what was observed for the mass on a spring, the position of the pendulum bob measured along the arc relative to its rest position is a function of the sine of the time.
Now suppose that we use our motion detector to investigate the how the velocity of the pendulum changes with respect to the time. As the pendulum bob does the back and forth , the velocity is continuously changing. There will be times at which the velocity is a negative value for moving leftward and other times at which it will be a positive value for moving rightward. If the variations in velocity over the course of time were plotted, the resulting graph would resemble the one shown below.
Now let's try to understand the relationship between the position of the bob along the arc of its motion and the velocity with which it moves. Suppose we identify several locations along the arc and then relate these positions to the velocity of the pendulum bob. The graphic below shows an effort to make such a connection between position and velocity. As is often said, a picture is worth a thousand words. Now here come the words. The plot above is based upon the equilibrium position D being designated as the zero position.
A displacement to the left of the equilibrium position is regarded as a negative position. A displacement to the right is regarded as a positive position. An analysis of the plots shows that the velocity is least when the displacement is greatest. And the velocity is greatest when the displacement of the bob is least. The further the bob has moved away from the equilibrium position, the slower it moves; and the closer the bob is to the equilibrium position, the faster it moves.
This can be explained by the fact that as the bob moves away from the equilibrium position, there is a restoring force that opposes its motion. Active 7 years, 2 months ago. Viewed 17k times. Improve this question. The acceleration can be resolved into two components in polar coordinates. One is radial and the other is tangential. I am just explaining to OP what elementary school textbooks mean.
My comment and Mike's comment are basically same. Add a comment. Active Oldest Votes. Improve this answer.
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