More Rotational Problems
NOTE: These solutions are not 100% proper. Vector signs are missing, FBDs are absent, some steps have been skipped. Use only as a guide.
1. Two country roads meet at an intersection. Each road is 7.5m wide. On a winter day, the coefficient of friction between the road and a car’s tires is 0.45. What is the highest speed a car can travel while making a left turn at the intersection? Would this be the same if the car was making a right turn?
The blue circle is to show that the turn is circular. The radius of the turn is the width of the road. Assume the car hugs the right side so R = 7.5m.
Left is the positive direction
Fnet = ma
Ff + Fg + Fn = ma
Ff = mac
µFn = mv2/R
µmg = mv2/R
0.45(9.81) = v2 / 7.5
v = 5.75 m/s
v = 5.8m/s = 21 km/h The fastest the car can make the turn at is 21km/h.
This would be very different if making a right turn. With a right turn, the turning radius is much smaller (blue circle). This makes Fc too large to handle, meaning the car would have to reduce speed v to make Fc back down to a smaller number (small enough for friction to handle). To make a right turn, a car needs to slow down almost to a stop before making the turn, or it will skid.
2. A car is traveling on an icy road that curves with a radius of 38m. The road is banked at an angle of 18.0 degrees. How fast can the car travel without flying off the road?
Assuming no friction.
Fnet = ma
Fn + Fg = mac
Fn + F1 + F2 = mac
F2 = mv2/R
mgtanθ = mv2/R
v = √gRtanθ
v = √ [9.81(38) tan (18)]
v = 11m/s or 40km/h
3. At an air show, a pilot pulls her plane out of a dive while going 210m/s. While making this turn, the pilot experiences 6.5g of acceleration. What is the radius of the turn?
The plane makes a vertical loop. The maximum Gee forces occur at the bottom of the loop. At this point, Flift is up and Fg is down.
Gee forces = Flift/Fg = 6.5 so let's solve for Flift = 6.5Fg = 6.5mg
Make up positive (toward the centre of the circle).
Fnet = ma
Flift + Fg = mac
6.5mg - mg = mv2/R *note mv2/R is positive because we said + is up.
5.5mg = mv2/R
R = v2/5.5g
R = 2102/(5.5*9.81)
R = 817m
The radius of the turn is 817m.
4. A woman on a bus carrying a 1.4kg purse notices the purse swings back at an angle of 7.0o from the vertical when the bus accelerates. Find the acceleration of the bus.
Purse swings back, bus accerlates forward. Forward (left in this diagram) is positive
Fnet = mac
T + Fg = mac
F1 + F2 + Fg = mac (vectors)
F2 = mac (not vectors any more)
mgtanθ = mac
ac = gtanθ
a = 9.81 tan(7.0)
a = 1.20 m/s2 The bus accelerates forward at 1.2 m/s2.
5. A woman on a bus carrying a 1.4kg purse notices the purse swings to the right at an angle of 12.0o from the vertical when the bus turns. Find the acceleration of the bus.
Same as #4 except different numbers and direction.
Purse swings to the right so bus must be turning to the left.
a = 9.81 tan(12.0) = 2.085 m/s2
The bus accelerates at 2.09m/s2 [left].
6. Little Jesse plays with his toy car set. He lays track along the floor and then builds a loop-the-loop that is 22cm tall. How fast must he push his toy car along the floor to allow it to successfully complete the loop?
At top: (down is positive)
Fnet = ma
Fn + Fg = mac
0 + mg = mv2/R (notice both Fg and ac are positive)
v = √gR
v = √9.81*0.11
v = 1.03 m/s (speed at top)
At bottom,
Etop = Ebottom
mgh + 1/2 mv2 = 1/2 mv2
v (bottom) = √ (2gh + v2 )
v (bottom) = √ 2 (9.81) ( 0.22) + 1.032
v (bottom = 2.31 m/s
He must push the car at least at 2.3m/s to make the loop.
7. Ahmed is taking the subway in Montreal. He stands in the middle of the subway car and does not hold on to the handrails. The subway train is moving at about 70km/h when it makes a smooth left turn of radius 300m. Ahmed leans to keep his balance. What is the angle of his lean?
See #4 for a similar FBD except Fn instead of T.
v = 70km/h = 19.4m/s
Fnet = ma
Fn + Fg = mac
Fn + F1 + F2 = mac
F2 = mv2/R
mgtanθ = mv2/R
tanθ = v2/gR
θ = tan-1( v2/gR)
θ = tan-1( 19.42/9.81*300)
θ = tan-1(0.127)
θ = 7.28o
He leans about 7.3 degrees into the turn (left).
8. David and Amanda are playing tetherball. They stand on opposite sides of the pole as shown. Amanda hits the ball and it travels in a horizontal circle at constant speed. Once the ball is struck, how long does David have to react before it reaches him?
R = 1.2 sin (40) = 0.771 m
Fnet = ma
T + Fg = mac
T + F1 + F2 = mac
F2 = mv2/R
mgtanθ = mv2/R
v = √gRtanθ
v = √ [9.81(0.771) tan (40)]
v = 2.51 m/s
David is 1/2 a circle away:
v = Δd/Δt
Δt = Δd / v
Δt = 2πR / v
Δt = 2π(0.771) / 2.51
Δt = 1.93s David has 1.9 seconds to react. Plenty of time for a coffee and then he can hit the ball.