All High School Physics Resources
Example Questions
Example Question #201 : Motion And Mechanics
A ball rolls around the edge of a circle with a radius of . If it has a centripial force of acting upon it, what is its speed?
Centripetal force is the force that constantly moves the object towards the center; it is what keeps the object moving in a circle rather than flying off tangentially to the circle.
The formula for force is .
Since we know the mass and the force, we can find the accleration.
Centripetal acceleration is given by the formula , where is the perceived tangential velocity and is the radius of the circle.
Plug in the given values and solve for the velocity.
Example Question #331 : Motion And Mechanics
A child spins a top with a radius of with a force of . How much torque is generated at the edge of the top?
Torque is a force times the radius of the circle, given by the formula:
In this case, we are given the radius in centimeters, so be sure to convert to meters:
Use this radius and the given force to solve for the torque.
Example Question #1 : Understanding Torque
Two children are trying to balance on a see-saw. One child has a mass of the other has a mass of . If the see-saw is balanced perfectly in the middle and the child is sitting at one end of the see-saw, how far from the center should the child sit so that the system is perfectly balanced?
If the see-saw is in total, then it has on either side of the fulcrum.
The question is asking us to find the equilibrium point; that means we want the net torque to equal zero.
Now, find the torque for the first child.
We are going to use the force of gravity for the force of the child.
When thinking of torque, treat the positive/negative as being clockwise vs. counter-clockwise instead of up vs. down. In this case, child one is generating counter-clockwise torque. That means that since , child two will be generating clockwise torque.
Solve for the radius (distance) of the second child.
Example Question #3 : Understanding Torque
A force is applied perpendicularly to a wrench. How much torque is generated?
The formula for torque is:
In this formula, is the angle the force makes with the lever arm. Since our force is applied perpendicularly, this angle will be . Use this angle, the force applied, and the length of the lever arm to calculate the torque.
Example Question #61 : Circular Motion
In ancient Greece, construction workers would use long levers to lift up boulders. Assume a boulder has a mass of and a certain construction worker has a mass of . The pivot point of the lever is placed from the boulder and the construction worker stands on the opposite end. How long does the lever need to be between the pivot point and the construction worker in order for the system to be in equilibrium?
We know that the system will be in equilibrium at the final point, meaning that the final net torque must be zero. That means that the torque generated by the boulder and the torque generated by the construction worker should be equal.
The negative sign is due to the placement of the two masses on opposite sides of the fulcrum. Essentially, one mass will have a positive radius and the other will have a negative radius. Use the definition of torque to expand this equation.
In this case, the forces will be the weights of the two objects due to gravity.
We can cancel the acceleration due to gravity from the equation to simplify.
Using the given information from the question, we can solve for the distance between the worker and the fulcrum. We know the mass of the worker, the mass of the boulder, and the distance between the boulder and the fulcrum.
Example Question #1 : Circular Motion
A shop sign weighing hangs from the end of a uniform beam as shown. Find the horizontal and vertical forces exerted by the hinge on the beam at the wall.
This is a static equilibrium problem. In order for static equilibrium to be achieved, there are three things that must be true. First, the sum of the forces in the horizontal direction must all equal . Second, the sum of the forces in the vertical direction must all be equal to zero. Third, the torque around a fixed axis must equal .
Let us begin by summing up the forces in the vertical direction.
Then let us sum up the forces in the horizontal direction
Lastly let us analyze the torque, using the hinge as the axis point where is the length of the beam.
Looking at these three equations, the easiest to work with would be our torque equation as it is only missing one variable. If we are able to find the tension in the direction we would then be able to use trigonometry to determine the tension in the cable overall.
We can now substitute this value back into our vertical direction equation to determine the force on the hinge in the direction.
We can go back to our tension in our direction and use trigonometry to determine the force of the tension in direction.
Rearrange and solve for the tension in the direction.
We can now go back to our second equation and substitute this value in.
Example Question #2 : Circular Motion
A pulley has a bucket of weight hanging from the cord over a well. The pulley has a mass of and radius of . There is a frictional torque of at the axle. Assume the cord has negligible mass and does not stretch or slip on the pulley. Calculate the linear acceleration of the bucket.
First let us analyze the torque that is happening on the bucket. There is a torque from the friction on the pulley and there is a torque from the bucket pulling on the pulley.
Next let us analyze the forces involved. There is a tension force pulling up on the bucket and there is the force of gravity pulling down on the bucket.
We can rearrange this to find so that we can substitute it into our torque equation.
Now substitute this into our torque equation
We know that there is a relationship between acceleration and angular acceleration.
So we can substitute this into our equation so that only linear acceleration is present.
We also know that the moment of inertia of the system is equal to the moment of inertia of the pulley plus the bucket.
We can now substitute this into our equation.
We can now starting putting in our known variables to solve for the missing acceleration.
Example Question #61 : Circular Motion
The torque applied to a wrench is . If the force applied to the wrench is , how long is the wrench?
The formula for torque is:
We are given the total torque and the force applied. Using these values, we can solve for the length of the wrench.
Example Question #3 : Circular Motion
Two children are trying to balance on a see-saw. One child has a mass of the other has a mass of . If the see-saw is balanced perfectly in the middle and the child is sitting at one end of the see-saw, how far from the center should the child sit so that the system is perfectly balanced?
If the see-saw is in total, then it has on either side of the fulcrum.
The question is asking us to find the equilibrium point; that means we want the net torque to equal zero.
Now, find the torque for the first child.
We are going to use the force of gravity for the force of the child.
When thinking of torque, treat the positive/negative as being clockwise vs. counter-clockwise instead of up vs. down. In this case, child one is generating counter-clockwise torque. That means that since , child two will be generating clockwise torque.
Solve for the radius (distance) of the second child.
Example Question #4 : Circular Motion
Two equal forces are applied to a door at the doorknob. The first force is applied to the plane of the door. The second is applied perpendicular to the door. Which force exerts a greater torque?
Both exert equal non-zero torques
The first applied at a angle
Both exert zero torques
The second applied perpendicular to the door
The second applied perpendicular to the door
Torque is equal to the force applied perpendicular to a surface, multiplied by the radius or distance to the pivot point.
In this case, one force is applied perpendicular and the other at an angle. The one that is applied at an angle, only has a small component of the total force acting in the perpendicular direction. This component will be smaller than the overall force. Therefore the force that is already acting perpendicular to door will provide the greatest torque.
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