Measure the length and the period
-- friction in the pivot -- air moving past the pendulum -- the effective length of the pendulum -- the local acceleration of gravity
-- its length (from the pivot to the center of mass of the swinging part) -- the local acceleration of gravity in the place where the pendulum is swinging
No, it does not. The earth's acceleration is relatively constant at or near the surface; about 9.8 meters per second squared. In short, just because the mass of an object is more or less does not mean it can affect the gravitational force of the earth. ================================= I think you may be asking whether the mass of the pendulum bob affects the result of the MEASUREMENT when we use that pendulum to measure the local acceleration of gravity. There again, the answer is No ... When you look at the formula that relates the period of the pendulum, its length, and the local gravity, the mass of the pendulum doesn't appear in the formula, and the result of the calculation is the same no matter how heavy your bob is. Now, if you want to get technical about it, the 'length' of the pendulum is the distance from the pivot to the center of mass. So, if the string or other means of suspension from which the bob hangs is NOT massless, then the mass of the bob does affect the position of the center of mass, and therefore the period of the pendulum. So for accurate measurement, it's always best to use the lightest possible string, and the most massive possible bob, in order to have the center of mass actually located as close as possible to where you THINK it is.
What you want is a pendulum with a frequency of 1/2 Hz. It swings left for 1 second,then right for 1 second, ticks once in each direction, and completes its cycle in exactly2 seconds.The length of such a pendulum technically depends on the acceleration due to gravityin the place where it's swinging. In fact, pendulum arrangements are used to measurethe local value of gravity.A good representative value for the length of the "seconds pendulum" is 0.994 meter.
Assuming the pendulum referred to s asimple pendulum of an arm and a weight the major factors on the period are the local attraction of gravity and the length of the arm.
-- friction in the pivot -- air moving past the pendulum -- the effective length of the pendulum -- the local acceleration of gravity
-- its length (from the pivot to the center of mass of the swinging part) -- the local acceleration of gravity in the place where the pendulum is swinging
T=2π√(L/g)where L is the length of the pendulum and g is the local acceleration of gravity.
Buoyancy and pressure determine whether the object floats or sinks.
No, it does not. The earth's acceleration is relatively constant at or near the surface; about 9.8 meters per second squared. In short, just because the mass of an object is more or less does not mean it can affect the gravitational force of the earth. ================================= I think you may be asking whether the mass of the pendulum bob affects the result of the MEASUREMENT when we use that pendulum to measure the local acceleration of gravity. There again, the answer is No ... When you look at the formula that relates the period of the pendulum, its length, and the local gravity, the mass of the pendulum doesn't appear in the formula, and the result of the calculation is the same no matter how heavy your bob is. Now, if you want to get technical about it, the 'length' of the pendulum is the distance from the pivot to the center of mass. So, if the string or other means of suspension from which the bob hangs is NOT massless, then the mass of the bob does affect the position of the center of mass, and therefore the period of the pendulum. So for accurate measurement, it's always best to use the lightest possible string, and the most massive possible bob, in order to have the center of mass actually located as close as possible to where you THINK it is.
What you want is a pendulum with a frequency of 1/2 Hz. It swings left for 1 second,then right for 1 second, ticks once in each direction, and completes its cycle in exactly2 seconds.The length of such a pendulum technically depends on the acceleration due to gravityin the place where it's swinging. In fact, pendulum arrangements are used to measurethe local value of gravity.A good representative value for the length of the "seconds pendulum" is 0.994 meter.
No; for example, other planets have more or less gravitational attraction than Earth. Even on Earth, there are local variations, so the standard acceleration of 9.8 or 9.82 meters/second squared is basically an average, or typical, value.
The period of a simple pendulum is given by the formulaT = 2*pi*sqrt(L/g)where T = periodL = lengthand g = local acceleration due to gravity.Note that this formula is applicable only when the angular displacement of the pendulum is small. For a displacement of 22.5 degrees (a quarter of a right angle), the true period is approx 1% longer : a clock will lose more than 1/2 a minute every hour!
32 grams I know this because of my scientific background I have also been to the moon.
Basically it is the object's "weight". The gravitational force on an object is its Mass X Gravitational Constant. The gravitational constant is the acceleration of a free falling body towards another body, and on Earth is equal to 9.81 meters/sec2 or 32.2 feet/sec2. Thus while the MASS of an object is a constant physical property, the WEIGHT of an object depends on the local gravity field pulling on that MASS.
Assuming the pendulum referred to s asimple pendulum of an arm and a weight the major factors on the period are the local attraction of gravity and the length of the arm.
It depends on the mass of the object, the local value of acceleration of gravity, and the object's height above the elevation you're using for your zero-potential-energy reference level.