W05D2 Concept Questions
Which of the following types of forces can produce a centripetal acceleration?
1.
2.
3.
4.
5.
Normal force
Frictional force
Tension force
Gravitational force
All of the above
Answer: 5. All of the above forces can produce a centripetal acceleration.
A normal force producing a centripetal acceleration could be the force of a car door on a
person while the car is turning, or the force on the person in the next question.
A friction force producing a centripetal acceleration could be the force on a person
standing on a rotating platform, or an object on a turntable (pre-CD, pre-I pod).
A tension force producing a centripetal acceleration could be an object whirling on the
end of a string; we have seen and will see many such examples.
Gravitational forces produce centripetal forces all the time; the moon orbiting the earth,
the earth or any planet orbiting the sun, … too many examples to list.
The “Barrel of Fun” must spin at a certain minimum angular speed in order for a rider
of mass m1 to “stick” to the wall. Does this minimum angular speed change for a rider of
mass m2  m1 ?
1. Yes
2. No
3. Not enough information to determine.
Answer: 2. The normal component N of the contact force that the cylinder wall exerts on
the rider pushes the rider toward the center of the orbit and is equal to N  m  2 R , where
m is the mass of the ride,  is the angular velocity and R is the radius of the barrel. The
vertical component of the contact force (static friction) supports the rider and hence is
equal to its maximum value, fs  mg . The normal force must be large enough in
magnitude to result in this frictional force, s N  s m  2 R  mg , or   g / Rs ,
independent of the mass of the rider. Note that in the limit s  0 , the necessary angular
speed becomes unboundedly large.
A stone attached to a string is whirled in a vertical plane. Let T1 , T2 , T3 and T4 be the
tensions at locations 1, 2, 3, and 4 required for the stone to have a given speed v0 at these
four locations. What are the relative tensions in the string?
1. T3  T2  T1  T4
2. T1  T2  T3  T4
3. T1  T2  T4  T3
4. None of the above
Answer: 3. When the string is in positions 2 and 4, the tension force is directed radially
inward and the tension is equal to T2 = T4  mv0 2 / R , where m is the mass of the stone
and R is the radius of the circle. When the string is in position 3 the tension and the
gravitational force points downward (inward), hence T3  mg  mv0 2 / R and thus
T3  mv0 2 / R  mg  T2  T4 . When the string is in position 1, the tension points upward
and the gravitational force points downward (outward), hence T3  mg  mv0 2 / R and thus
T3  mv0 2 / R  mg  T2  T4 .
A puck of mass M is moving in a circle at uniform speed on a frictionless table as
shown below. The puck is attached to a massless, frictionless string that passes through a
hole in the table and which is in turn attached to a suspended bob, also of mass M , at
rest below the table. What is the magnitude of the centripetal acceleration of the moving
puck?
1.
2.
3.
4.
5.
Less than g
Equal to g
Greater than g
Zero
Insufficient information
Answer: 2. Since the suspended bob is static, the tension in string is equal to the
gravitational force acting on the bob, T  Mg . The puck is accelerating inward due to the
tensile force of the string acting on the puck, hence T  Ma . Thus Mg  Ma , or g  a .