Ch 5.1 Help
Page 1 of 1 • Share •
Ch 5.1 Help
Question 1: (1 point)
An angle subtends a fraction 0.1146 of the circumference of a circle. What is the angle in radians (see sheet 3) ? Indicate with a negative (positive) sign whether your numerical result when expressed in degrees is smaller (larger) than your result in radians.
How to solve:
Question 2: (1 point)
An object moves with constant speed on a circle of 0.5314 m radius due to a centripetal acceleration of 4.306 m/s2. What is the speed of the object (see sheet 4) ? Indicate with a negative (positive) sign whether the tangential acceleration is zero (nonzero) (see sheet 7).
How to solve:
Question 3: (1 point)
An object moving on a circle with radius of 0.4736 m makes 36.2 revolutions per minute. What is the centripetal acceleration (see sheet 4) ? Indicate with a negative (positive) sign whether the period can be (cannot be) calculated from the input given.
How to solve:
Question 4: (1 point)
You swing an 0.5156 kg mass held by a string in a horizontal circle of 0.6398 m radius making a turn every 2.1644 seconds. What is the centripetal force (see sheet 4,10) ? Indicate with a positive (negative) sign whether the centripetal force is directed away from (toward) your hand holding the string.
How to solve:
An angle subtends a fraction 0.1146 of the circumference of a circle. What is the angle in radians (see sheet 3) ? Indicate with a negative (positive) sign whether your numerical result when expressed in degrees is smaller (larger) than your result in radians.
How to solve:
Question 2: (1 point)
An object moves with constant speed on a circle of 0.5314 m radius due to a centripetal acceleration of 4.306 m/s2. What is the speed of the object (see sheet 4) ? Indicate with a negative (positive) sign whether the tangential acceleration is zero (nonzero) (see sheet 7).
How to solve:
Question 3: (1 point)
An object moving on a circle with radius of 0.4736 m makes 36.2 revolutions per minute. What is the centripetal acceleration (see sheet 4) ? Indicate with a negative (positive) sign whether the period can be (cannot be) calculated from the input given.
How to solve:
Question 4: (1 point)
You swing an 0.5156 kg mass held by a string in a horizontal circle of 0.6398 m radius making a turn every 2.1644 seconds. What is the centripetal force (see sheet 4,10) ? Indicate with a positive (negative) sign whether the centripetal force is directed away from (toward) your hand holding the string.
How to solve:
Guest10 Guest
Question 1
I hope this makes sense. I'm trying to embed the image into this post, if that doesn't work just click the link for a drawing of Question 1.
Guest10 Guest
Re: Ch 5.1 Help
Ok...so embedding images doesn't work. Copy the link below below for a question 1 drawing.
Guest10 Guest
Re: Ch 5.1 Help
That didn't work either...here we go.
[]http://img157.imageshack.us/my.php?image=newbitmapimagenf6.png
[]http://img157.imageshack.us/my.php?image=newbitmapimagenf6.png
Guest10 Guest
Re: Ch 5.1 Help
Question 1: (Answer is positive)
An angle subtends a fraction 0.1146 of the circumference of a circle. What is the angle in radians (see sheet 3) ? Indicate with a negative (positive) sign whether your numerical result when expressed in degrees is smaller (larger) than your result in radians.
How to solve:
The value that the question gives you is the radius. And it is asking for the circumference. So use the equation: C = 2(pi)r
C = 2(pi)r
C = 2(.1146)(3.14)
C = .7201
An angle subtends a fraction 0.1146 of the circumference of a circle. What is the angle in radians (see sheet 3) ? Indicate with a negative (positive) sign whether your numerical result when expressed in degrees is smaller (larger) than your result in radians.
How to solve:
The value that the question gives you is the radius. And it is asking for the circumference. So use the equation: C = 2(pi)r
C = 2(pi)r
C = 2(.1146)(3.14)
C = .7201
ravenscorne Posts : 15
Join date : 20080920
Re: Ch 5.1 Help
Question 2: (Answer is negative)
An object moves with constant speed on a circle of 0.5314 m radius due to a centripetal acceleration of 4.306 m/s2. What is the speed of the object (see sheet 4) ? Indicate with a negative (positive) sign whether the tangential acceleration is zero (nonzero) (see sheet 7).
How to Solve:
To solve this problem, use the centripetal acceleration formula: a_c = v^2/r. Since we're given a_c and r, we just plug in to solve for v.
v = sqrt (a_c*r)
v = sqrt (4.306*0.5314)
v = 1.5127
An object moves with constant speed on a circle of 0.5314 m radius due to a centripetal acceleration of 4.306 m/s2. What is the speed of the object (see sheet 4) ? Indicate with a negative (positive) sign whether the tangential acceleration is zero (nonzero) (see sheet 7).
How to Solve:
To solve this problem, use the centripetal acceleration formula: a_c = v^2/r. Since we're given a_c and r, we just plug in to solve for v.
v = sqrt (a_c*r)
v = sqrt (4.306*0.5314)
v = 1.5127
ravenscorne Posts : 15
Join date : 20080920
Re: Ch 5.1 Help
Question 3: (Answer is negative)
An object moving on a circle with radius of 0.4736 m makes 36.2 revolutions per minute. What is the centripetal acceleration (see sheet 4) ? Indicate with a negative (positive) sign whether the period can be (cannot be) calculated from the input given.
How to solve:
To solve this question, we need to find the sentripetal acceleration. That means we need the radius, which we have and the velocity, which we need to find out.
To solve this, first start with the revolutions. If an object makes 36.2 revolutions in 1 minute, we know that it makes 36.2 revolutions in 60 seconds. So, 36.2/60 = .6033 revolutions per second. That means it travels this fraction of a circle in one second.
Now we find the distance that this fraction actually takes. So that means we need to find the circumference of the circle. C = 2(pi)r.
C = 2(3.14)(.4736)
C = 2.9757
So now in one second, .6033 of the circumference is 1.795.
This value, 1.795 is actually the velocity, since it takes one second to travel this amount of distance.
Now back to the first equation, because we now have both velocity and radius, we can solve for centripetal acceleration.
a_c = v^2/r
a_c = (1.795)^2/(.4736)
a_c = 6.805
An object moving on a circle with radius of 0.4736 m makes 36.2 revolutions per minute. What is the centripetal acceleration (see sheet 4) ? Indicate with a negative (positive) sign whether the period can be (cannot be) calculated from the input given.
How to solve:
To solve this question, we need to find the sentripetal acceleration. That means we need the radius, which we have and the velocity, which we need to find out.
To solve this, first start with the revolutions. If an object makes 36.2 revolutions in 1 minute, we know that it makes 36.2 revolutions in 60 seconds. So, 36.2/60 = .6033 revolutions per second. That means it travels this fraction of a circle in one second.
Now we find the distance that this fraction actually takes. So that means we need to find the circumference of the circle. C = 2(pi)r.
C = 2(3.14)(.4736)
C = 2.9757
So now in one second, .6033 of the circumference is 1.795.
This value, 1.795 is actually the velocity, since it takes one second to travel this amount of distance.
Now back to the first equation, because we now have both velocity and radius, we can solve for centripetal acceleration.
a_c = v^2/r
a_c = (1.795)^2/(.4736)
a_c = 6.805
ravenscorne Posts : 15
Join date : 20080920
Re: Ch 5.1 Help
Question 4: (Answer is negative)
You swing an 0.5156 kg mass held by a string in a horizontal circle of 0.6398 m radius making a turn every 2.1644 seconds. What is the centripetal force (see sheet 4,10) ? Indicate with a positive (negative) sign whether the centripetal force is directed away from (toward) your hand holding the string.
How to solve:
Centripetal force works the same way as regular force in that the equation is still F=m*a, except the acceleration in this instance is centripetal acceleration. For this, we have the mass, and we need to find the cetripetal acceleration.
To find centripetal acceleration, we need to use the equation a_c = V^2/r. And we have the radius, which means we need to find the velocity.
Now the equation says that you make a turn every 2.1664 seconds. So to find how much of the circle you travel in 1 second, we say:
1/(2.1644) = .4620 This value is the fraction of the circle that you travel in 1 second.
Now to find the circumference, C = 2(pi)r.
C = 2(3.14)(.6398)
C = 4.01998
Multiply this value by the fraction of the circle we found,
4.01998*0.4620 = 1.8572
This is now the velocity in meters per second. So we can use this value in the a_c equation, and thus the F=m*a to solve the problem.
a_c = v^2/r
a_c = (1.8572)^2/(.6398)
a_c = 5.3912
F_ac = m*a_c
F_ac = 0.5156*5.3912
F_ac = 2.7797
You swing an 0.5156 kg mass held by a string in a horizontal circle of 0.6398 m radius making a turn every 2.1644 seconds. What is the centripetal force (see sheet 4,10) ? Indicate with a positive (negative) sign whether the centripetal force is directed away from (toward) your hand holding the string.
How to solve:
Centripetal force works the same way as regular force in that the equation is still F=m*a, except the acceleration in this instance is centripetal acceleration. For this, we have the mass, and we need to find the cetripetal acceleration.
To find centripetal acceleration, we need to use the equation a_c = V^2/r. And we have the radius, which means we need to find the velocity.
Now the equation says that you make a turn every 2.1664 seconds. So to find how much of the circle you travel in 1 second, we say:
1/(2.1644) = .4620 This value is the fraction of the circle that you travel in 1 second.
Now to find the circumference, C = 2(pi)r.
C = 2(3.14)(.6398)
C = 4.01998
Multiply this value by the fraction of the circle we found,
4.01998*0.4620 = 1.8572
This is now the velocity in meters per second. So we can use this value in the a_c equation, and thus the F=m*a to solve the problem.
a_c = v^2/r
a_c = (1.8572)^2/(.6398)
a_c = 5.3912
F_ac = m*a_c
F_ac = 0.5156*5.3912
F_ac = 2.7797
ravenscorne Posts : 15
Join date : 20080920
Page 1 of 1
Permissions in this forum:
You cannot reply to topics in this forum

