Physics 1307 Practice Quiz 5 Chapter 5 Solutions
Physics 1307 Practice Quiz 5 Chapter 5 Problem 1 A rollercoaster vehicle has a mass of 500 kg when fully loaded with passengers. (a) If the vehicle has a speed of 20 m/s at point a, what is the force of the track of the vehicle at this point? (b) What is the maximum speed the vehicle can have at point b for gravity to hold it in the track?
b + 15 m 10 m
a Problem 2 Io a satellite of Jupiter has an orbital period of 1.77 days and an orbital radius of 4.22 ×10 5 km . From these data, determine the mass of Jupiter.
Solutions (a) Let us analyze the forces on the roller coaster vehicle at point a:
FN
mg 1/2
Observing the figure we can deduce that the cart is moving in circular motion, thus: F N − mg = m
v2
where we R have assigned a positive sign to the centripetal acceleration since points upward. Remember the centripetal force is not an extra force is just Newton's second law applied to circular motion. The force of the track is equal to the normal force so we get : F N =mgm
v2 R
=500 kg
[
20 m/ s 2 10 m
2
]
9.8 m/ s =24900 N
(b) The forces at b are quite similar, the only difference is the direction of the centripetal acceleration that this time v2 F −mg=−m is negative so: . The maximum speed is reached when the normal force is zero (why?) N R therefore Rg=v=12 m/s
Problem 2 Kepler's third law states that
r3 T2
=constant . The value of this constant can be deduced using Newton's laws
(see textbook page 203) and it is equal to r3 2
M=
r 3 4 2 GT
2
=
=
GM 4
2
, where M is the mass of the central body. So we are ready !
GM
T 4 2 4.22×108 m3 4 2 7×10 −11 N m2 / kg2 [ 1.77 days86400 s/1 day ]
2/2
2
=1.81×10 27 kg
b + 15 m 10 m
a Problem 2 Io a satellite of Jupiter has an orbital period of 1.77 days and an orbital radius of 4.22 ×10 5 km . From these data, determine the mass of Jupiter.
Solutions (a) Let us analyze the forces on the roller coaster vehicle at point a:
FN
mg 1/2
Observing the figure we can deduce that the cart is moving in circular motion, thus: F N − mg = m
v2
where we R have assigned a positive sign to the centripetal acceleration since points upward. Remember the centripetal force is not an extra force is just Newton's second law applied to circular motion. The force of the track is equal to the normal force so we get : F N =mgm
v2 R
=500 kg
[
20 m/ s 2 10 m
2
]
9.8 m/ s =24900 N
(b) The forces at b are quite similar, the only difference is the direction of the centripetal acceleration that this time v2 F −mg=−m is negative so: . The maximum speed is reached when the normal force is zero (why?) N R therefore Rg=v=12 m/s
Problem 2 Kepler's third law states that
r3 T2
=constant . The value of this constant can be deduced using Newton's laws
(see textbook page 203) and it is equal to r3 2
M=
r 3 4 2 GT
2
=
=
GM 4
2
, where M is the mass of the central body. So we are ready !
GM
T 4 2 4.22×108 m3 4 2 7×10 −11 N m2 / kg2 [ 1.77 days86400 s/1 day ]
2/2
2
=1.81×10 27 kg
