Problem#1
A light string can support a stationary hanging load of 25.0 kg before breaking. A 3.00-kg object attached to the string rotates on a horizontal, frictionless table in a circle of radius 0.800 m, while the other end of the string is held fixed. What range of speeds can the object have before the string breaks?
Answer:
Given: m = 3.00 kg , r = 0.800 m.
The string will break if the tension exceeds the weight corresponding to 25.0 kg, so
Tmax = Mg = 25.0 kg(9.80 m/s2) = 245 N
When the 3.00 kg mass rotates in a horizontal circle, the tension causes the centripetal acceleration, so
T = mv2/r
v2 = Tr/m ≤ rTmax/m = (0.800 m)(245 N)/3.00 kg = 65.3 m2/s2
and
0 ≤ v ≤ 8.08 m/s
Problem#2
A curve in a road forms part of a horizontal circle. As a car goes around it at constant speed 14.0 m/s, the total force on the driver has magnitude 130 N. What is the total vector force on the driver if the speed is 18.0 m/s instead?
Answer:
In ∑F = mv2/r, both m and r are unknown but remain constant. Therefore, ∑F is proportional to v2 and increases by a factor of
(18.0/14.0)2
as v increases from 14.0 m/s to 18.0 m/s. The total force at the higher speed is then
∑Ffast = (18.0/14.0)2(130 N) = 215 N
Symbolically, write
∑Fslow = (m/r)(14.0 m/s)2 and
∑Ffast = (m/r)(18.0 m/s)2
Dividing gives
∑Ffast/∑Fslow = (18.0/14.0)2
∑Ffast = (18.0/14.0)2∑Fslow = (18.0/14.0)2(130 N) = 215 N
This force must be horizontally inward to produce the driver’s centripetal acceleration.
Problem#3
In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.20 x 106 m/s.
Find (a) the force acting on the electron as it revolves in a circular orbit of radius 0.530 x 10-10 m and (b) the centripetal acceleration of the electron.
Answer:
(a) F = mv2/r
F = (9.11 x 10-31 kg)(2.20 x 106 m/s)2/(0.530 x 10-10 m) = 8.32 x 10-8 N inward
(b) a = F/m = 8.32 x 10-8 N inward/(9.11 x 10-31 kg) = 9.13 x 1022 m/s2 inward
Problem#4
In a cyclotron (one type of particle accelerator), a deuteron (of atomic mass 2.00 u) reaches a final speed of 10.0% of the speed of light while moving in a circular path of radius 0.480 m. The deuteron is maintained in the circular path by a magnetic force. What magnitude of force is required?
Answer:
F = mv2/r
F = (2 x 1.661 x 10-27 kg)(10% x 2.998 x 108 m/s)2/(0.480 m) = 6.22 x 10-12 N
Problem#5
A coin placed 30.0 cm from the center of a rotating, horizontal turntable slips when its speed is 50.0 cm/s. (a) What force causes the centripetal acceleration when the coin is stationary relative to the turntable? (b) What is the coefficient of static friction between coin and turntable?
Answer:
(a) force causes the centripetal acceleration when the coin is stationary relative to the turntable is static friction.
(b) ∑Fy = may
n – mg = 0
n = mg
and
∑Fx = max
f = mv2/r
µmg = mv2/r
µ = v2/gr = (50.0 cm/s)2/[(980 cm/s2)(30.0 cm)] = 0.0850
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