Cambridge IGCSE Physics Moments Revision Notes PDF

Head to www.savemyexams.com for more awesome resources Cambridge (CIE) IGCSE Physics Your notes Moments Contents Moments Equilibrium Centre of Gravity Investigating Centre of Gravity...

Head to www.savemyexams.com for more awesome resources Cambridge (CIE) IGCSE Physics Your notes Moments Contents Moments Equilibrium Centre of Gravity Investigating Centre of Gravity Page 1 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Moments Your notes Moments The moment of a force is the turning effect produced when a force is exerted on an object Examples of the turning effect of a force are: A child on a see-saw Turning the handle of a spanner A door opening and closing Using a crane to move building supplies Using a screwdriver to open a tin of paint Turning a tap on and off Picking up a wheelbarrow Using scissors Forces can cause the rotation of an object about a fixed pivot This rotation can be clockwise or anticlockwise Clockwise and anti-clockwise rotation Page 2 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Consider the hands of a clock when deciding if an object will rotate in a clockwise or anti-clockwise direction A force applied on one side of the pivot will cause the object to rotate Turning effect of a force about a pivot Page 3 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes The force applied will cause the object to rotate clockwise about the pivot The moment equation A moment is defined as: The turning effect of a force about a pivot The size of a moment is defined by the equation: moment = force × perpendicular distance from pivot The forces should be perpendicular to the distance from the pivot For example, on a horizontal beam, the forces which will cause a moment are those directed upwards or downwards The turning effect of a force exerted on a spanner Page 4 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes The moment depends on the force and perpendicular distance to the pivot Increasing the distance at which a force is applied from a pivot decreases the force required If you try to push open a door right next to the hinge, it is very difficult, as it requires a lot of force If you push the door open at the side furthest from the hinge, then it is much easier, as less force is required Forces required to open a door Page 5 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes A greater force is required to push open a door next to the hinges than at the door handle Worked Example A carpenter attempts to loosen a bolt that has rusted. To turn the bolt, they exert a force of 22 N using a spanner of length 20 cm. The force is exerted 5 cm from the end of the spanner. Calculate the turning effect of the force. Answer: Step 1: List the known quantities Force, F = 22 N Length of spanner, = 20 cm Step 2: Determine the distance from the pivot The force is exerted 5 cm from the end of the spanner Therefore, the distance from the force to the pivot is s = 20 − 5 Page 6 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources s = 15 cm Convert cm to m Your notes 15 s= 100 s = 0. 15 m Step 3: Write out the equation for moments moment = force × perpendicular distance from pivot M = Fs Step 4: Substitute in the known values to calculate M = 22 × 0. 15 M = 3.3 N m Examiner Tips and Tricks The moment of a force is measured in newton metres (N m), but can also be newton centimetres (N cm) if the distance is measured in cm instead. If your IGCSE moments exam question doesn't ask for a specific unit, always convert the distance into metres Principle of moments (core) The principle of moments states that: If an object is balanced, the total clockwise moment about a pivot equals the total anticlockwise moment about that pivot The principle of moments means that for a balanced object, the moments on both sides of the pivot are equal clockwise moment = anticlockwise moment Principle of moments Page 7 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Imagine holding the beam about the pivot and applying just one of the forces. If the beam moves clockwise then the force applied is clockwise. This is the principle of moments Worked Example A parent and child are at opposite ends of a playground see-saw. The weight force acting on the parent is 690 N and the weight force acting on the child is 140 N. The adult sits at a distance of 0.3 m from the pivot. Page 8 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Calculate the distance the child must sit from the pivot for the see-saw to be balanced. Use the principle of moments in your calculation. Answer: Step 1: List the known quantities Clockwise force (child), F = 140 N child Anticlockwise force (adult), F = 690 N adult Distance of adult from the pivot, s = 0.3 m adult Step 2: Write down the moment equation and the principle of moments Moment equation: moment = force × perpendicular distance from pivot M = Fs Principle of moments: total clockwise moments = total anticlockwise moments Step 3: Calculate the total clockwise moments The clockwise moment is from the child M = child F child × s child Page 9 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources M child = 140 × s child Step 4: Calculate the total anticlockwise moments Your notes The anticlockwise moment is from the adult M =F adult adult × s adult M adult = 690 × 0. 3 M adult = 207 N m Step 5: Substitute into the principle of moments equation total clockwise moments = total anticlockwise moments M child = M adult 140 × s child = 207 Step 6: Rearrange for the distance of the child from the pivot 207 s child = 140 s child = 1. 5 m The child must sit 1.5 m from the pivot to balance the see-saw Examiner Tips and Tricks Make sure that all the distances are in the same units and that you’re considering the correct forces as clockwise or anticlockwise If you are studying the core tier for IGCSE Physics, you will only be expected to apply the principle of moments to a situation where one force acts on either side of the pivot Principle of moments (extended) Extended tier only Page 10 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources In the example below, the forces and distances of the objects on the beam are different, but they are arranged in a way that balances the whole system Your notes Using the principle of moments The clockwise and anticlockwise moments acting on a beam are balanced In the above diagram: Force F causes an anticlockwise moment of F × d about the pivot 1 1 1 Force F causes a clockwise moment of F × d about the pivot 2 2 2 Force F causes an anticlockwise moment of F × d about the pivot 3 3 3 Collecting the clockwise and anticlockwise moments: Sum of the clockwise moments = F × d 2 2 Sum of the anticlockwise moments = (F × d ) + (F × d ) 1 1 3 3 Page 11 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Using the principle of moments, the beam is balanced when: sum of the clockwise moments = sum of the anticlockwise moments Your notes F 2 × d 2 = ( F 1 × d 1) + ( F 3 × d 3) Page 12 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Equilibrium Your notes Equilibrium In physics, the term equilibrium means: A state of balance or stability In other words, a system in equilibrium keeps doing what it’s doing without any change Conditions for equilibrium For objects in equilibrium: The forces on the object must be balanced There must be no resultant force The sum of clockwise moments on the object must equal the sum of anticlockwise moments There must be no resultant moment Examples of systems in equilibrium When the forces and moments on an object are balanced, the object will remain in equilibrium Demonstrating equilibrium Page 13 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Extended tier only Aim of the experiment Your notes This experiment aims to demonstrate that there is no resultant moment for an object in equilibrium Variables Independent variable = force, F , and distance, s Dependent variable = moment, M Control variables: The length of the cotton loops should be equal on each side of the beam Equipment Equipment list Equipment Purpose Metre ruler with a small hole at the centre To provide the beam on which to add masses 2 × 100 g mass hangers To attach the masses to the ruler 8 × 100 g masses To add the mass at different points along the ruler Clamp stand, boss & clamp To secure the pivot in place Optical pin and cork To act as the pivot Small piece of plasticine To ensure the ruler is balanced at the start 2 loops of cotton To attach the mass hangers to the metre ruler Example set up of equipment to demonstrate equilibrium Page 14 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes The ruler acts as the beam with the pin as the pivot. Unequal masses are added at different distances until the beam is balanced and equilibrium is reached Method 1. Hang unequal loads on either side of the pivot; one person holds the beam while the other person hangs the loads 2. Adjust the distances of mass 1, m , and mass 2, m , until the beam is balanced 1 2 3. Adjust further to ensure the beam is perfectly horizontal with no resultant moment 4. Record the distance from the pivot of masses m and m 1 2 5. Repeat the process for different sized loads Example results table Page 15 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes A results table should contain spaces for all the measurements taken and any calculations required Analysis of results Force 1, F , is providing the anticlockwise moment, M 1 1 Where: F 1 = m 1g M 1 = F 1s 1 Force 2, F , is providing the clockwise moment, M 2 2 Where: F 2 = m 2g M 2 = F 2s 2 Remember to convert g to kg and cm to m for the calculations to give units of Nm for the moments The results should show that for all the systems tested, the anticlockwise moment is equal to the clockwise moment Therefore, there is no resultant moment when the system is in equilibrium Page 16 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Evaluating the experiment Systematic errors Your notes The cotton loops should be added to the ruler when viewed straight on to avoid a parallax error The cotton loops should be measured to ensure they are equal in length The experiment should be checked to ensure there is no friction between the metre ruler and the optical pin pivot so the ruler is balanced, only because of the added masses Random errors The precision of the experiment is improved by: ensuring the experiment is done in a space with no draft or breeze, as this could affect the motion or position of the hanging masses using an electronic system or a spirit level that identifies the angle of the beam would improve the experiment, or using a flat rod with masses placed on top The accuracy of the experiment is improved by: taking more than five readings for each mass and position and then calculating the mean It is assumed that the mass of the cotton loops is negligible (zero) It is assumed that the mass of each mass and hangar is 100 g, this should be verified in advance using an electronic balance Safety considerations Safety goggles should be worn because the cotton loops could snap and hit someone in the eye Use a G clamp to secure the clamp stand to the bench so it does not topple over and cause injury Stand up to carry out this experiment so you do not fall over when looking level with the metre ruler Place a mat or a soft material below the metre ruler to cushion any masses that may fall to the ground and to keep the area clear of feet and hands Page 17 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Centre of Gravity Your notes Centre of gravity The centre of gravity of an object is defined as: The point through which the weight of an object acts For a symmetrical object of uniform density, the centre of gravity is located at the point of symmetry For example, the centre of gravity of a sphere is at the centre Finding the centre of gravity of symmetrical objects Page 18 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources The centre of gravity of a regular shaped object can be found using symmetry Your notes Examiner Tips and Tricks Since the centre of gravity is a hypothetical point, it can lie inside or outside of a body. The centre of gravity will constantly shift depending on the shape of a body. For example, a human body’s centre of gravity is lower when learning forward than when standing upright However, when you are drawing force diagrams, always draw the weight force as if it were acting from the centre of gravity of the object! Stability The centre of gravity of a symmetrical object is along the axis of symmetry The position of the centre of gravity affects the stability of an object An object is stable when its centre of gravity lies above its base Centre of gravity of an object in different positions The object will topple, when its centre of gravity is no longer over its base Page 19 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources If the centre of gravity does not lie above its base, then an object will topple over Your notes The most stable objects have a low centre of gravity and a wide base Low centre of gravity of a car Cars are stable because they have a low centre of gravity and a wide base Taller objects with a narrow base have a higher centre of gravity and are less stable This is why lorries and buses are advised not to use motorways and bridges on very windy days Stability of object in different positions When the object is positioned on its narrow base, it is less stable because its centre of gravity is higher Moments and stability If the line of action of the weight force lies outside the base of the object, there will be a resultant moment, and the body will topple Car and bus on varying inclines Page 20 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes The car can be titled to 60° without toppling, but the bus will topple at 45° Tall objects with a narrow base will topple easily Ten-pin bowling pins are designed specifically to topple easily The stability of objects can be increased by widening the base High chairs are designed with a wide base so that they do not topple Bunsen burners have a wide base to ensure they do not topple Examiner Tips and Tricks Since the centre of gravity is a hypothetical point, it can lie inside or outside of a body. The centre of gravity will constantly shift depending on the shape of a body. For example, a human body’s centre of gravity is lower when learning forward than upright Page 21 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Investigating Centre of Gravity Your notes Investigating centre of gravity Aim of the experiment This experiment aims to determine the position of the centre of gravity of an irregularly shaped plane lamina Independent variable = position of centre of gravity Dependent variable = shape of plane lamina Control variables: Punching the holes in the plane lamina before determining the line of action of the weight force Equipment Equipment list Equipment Purpose Irregularly shaped plane lamina To find the position of the centre of gravity of Hole punch To create a hole in the plane lamina to tie the thread to Thread To hang the plane lamina from the clamp Thread and mass or sticky tack To form the plumb line Pencil and ruler To draw the line of action of the weight force Clamp stand, boss & clamp To hang the plane lamina from Method Determining the centre of gravity of an irregularly shaped plane lamina Page 22 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes For irregularly shaped objects, the centre of gravity can be found using the suspension method 1. Punch 3 holes near the outer edges of the plane lamina in different locations 2. Create a loop of thread and hang the plane lamina from the clamp 3. Use a plumb line (a weighted thread) aligned with the hanging thread to show the line of action of the weight force 4. Use a ruler and pencil to mark the line of action of the weight force onto the plane lamina 5. Repeat the process until 3 lines have been drawn 6. The point at which the lines cross is the position of the centre of gravity Analysis of results Page 23 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Each plane lamina is an irregularly shaped object When an object is suspended from a point, it will always settle with its centre of gravity directly below Your notes the point of suspension Evaluating the experiment Systematic errors The plumb line should be viewed straight on to avoid a parallax error Dots should be made on the plane lamina whilst in position and then a ruled line should be made after the lamina has been removed from the clamp Random errors The plane lamina should be allowed to settle before determining the action line of the weight force The holes should be punched in the plane lamina before determining the action line of the weight force, because punching the holes after can affect the position of the centre of gravity Page 24 of 24 © 2015-2024 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers