Date | November 2020 | Marks available | 3 | Reference code | 20N.2.SL.TZ0.2 |
Level | Standard level | Paper | Paper 2 | Time zone | 0 - no time zone |
Command term | Deduce | Question number | 2 | Adapted from | N/A |
Question
The Rotor is an amusement park ride that can be modelled as a vertical cylinder of inner radius RR rotating about its axis. When the cylinder rotates sufficiently fast, the floor drops out and the passengers stay motionless against the inner surface of the cylinder. The diagram shows a person taking the Rotor ride. The floor of the Rotor has been lowered away from the person.
Draw and label the free-body diagram for the person.
The person must not slide down the wall. Show that the minimum angular velocity ωω of the cylinder for this situation is
ω=√gμRω=√gμR
where μμ is the coefficient of static friction between the person and the cylinder.
The coefficient of static friction between the person and the cylinder is 0.400.40. The radius of the cylinder is 3.5 m3.5m. The cylinder makes 2828 revolutions per minute. Deduce whether the person will slide down the inner surface of the cylinder.
Markscheme
arrow downwards labelled weight/W/mgmg and arrow upwards labelled friction/FF ✓
arrow horizontally to the left labelled «normal» reaction/NN ✓
Ignore point of application of the forces but do not allow arrows that do not touch the object.
Do not allow horizontal force to be labelled ‘centripetal’ or RR.
See F=μNF=μN AND N=mRω2N=mRω2 ✓
«substituting for N» μmω2R=mgμmω2R=mg ✓
ALTERNATIVE 1
minimum required angular velocity «=√9.810.40×3.5»=2.6«rad s-1» ✓
actual angular velocity «=2π(6028)»=2.9«rad s-1»✓
actual angular velocity is greater than the minimum, so the person does not slide ✓
ALTERNATIVE 2
Minimum friction force =mg=«9.81 m» ✓
Actual friction force «=μmRω2=0.40 m×3.5(2π2860)2»=12.0 m ✓
Actual friction force is greater than the minimum frictional force so the person does not slide ✓
Allow 2.7 from g=10 ms-2.