ENPH 131
Last Name:
First Name:
and
Seminar 7
FBD/KD Problems 2 –
Normal/Tangential Coords.
Department of
Physics
Starting Mar 10, 2014
Faculty of
Engineering
ID Number:
Seminar Section:
Please do all work on handout and box final numerical answers. Total=20 marks.
1. [3] A small ball is attached to a vertical rod by two strings as shown. The system rotates
in a horizontal circle at a constant speed. Draw the FBD/KD diagrams for the ball.
Solution
T1
T2
=
man
mg
mat=0
mab=0
(2 marks for the FBD, 1 mark for the KD. Tensions must be labeled distinctly.)
2. [8] An 1200 kg car travels horizontally on a circular track of radius R = 120 m by
driving on a banked incline (θ = 36.87°) with friction. If the car travels at a constant 32.0
m/s without slipping up or down the incline, find the magnitude and direction of the static
frictional force acting on the car (as a whole). Neglect the size of the car, and for simplicity
use sin(θ) = 3/5, and cos(θ) = 4/5.
Solution
It is not obvious which direction fs acts, so we will choose a direction and let the numbers
decide. Assuming fs acts parallel and down the incline, the FBD/KD for the situation at the
instant shown is:
N
man
=
fs
mg
mat=0
mab=0
(2 marks for the FBD/ 1 for the KD: it’s fine if students have chosen fs to act up the incline.)
The two nontrivial equations of motion are (for fs as indicated):
∑ Fn = fs cosθ + N sinθ = mv 2 / R
(2 marks)
∑ Fb = N cosθ − fs sinθ − mg = 0
Solving the second for N yields: N = (mg + fs sinθ ) / cosθ , and inserting into the first:
fs (cosθ + sin 2 θ / cosθ ) + mg tan θ = mv 2 / R ⇔ fs / cosθ + mg tan θ = mv 2 / R
whence:
fs = mv 2 cosθ / R − mgsin θ = 1200(32 2 ⋅ 0.8 /120 − 9.81⋅ 0.6) ≈ 1.13kN .
Since this turned out to be numerically positive, the static frictional force at that speed acts
down the incline. (3 marks for the solution of the system for the numerical value of fs.)
3. [9] A small 2.00 kg collar is confined to move along a smooth rod in the vertical plane
described by
y(x) = x2/2
(with x and y in m). The collar is initially at x = 0, moving with a speed of 3.80 m/s in the
positive x direction. When it reaches x = 0.750 m, determine:
a) the rate of change of speed of the collar
b) the magnitude of the normal force the rod exerts on the collar.
(Note: in order to use at ds = v dv, you will need the following ‘infinitesimal’ triangle to
replace ds.)
ds
dy
θ
dx
Solution
a) The parabolic rod can be thought of as a variable angle incline, and so the FBD/KD at a
general ‘angle’ is:
N
n+
man
t+
=
θ
mat
mg
(2 marks for the FBD/KD)
The equation of motion for the tangential direction is:
∑ Ft = −mgsinθ = mat (1 mark)
so at = −g sinθ. (Note the minus sign is physical: the object is slowing down.)
For this curve: dy/dx = tanθ = x. So at x = 0.75, tanθ = 3/4, sinθ = 3/5 (and cosθ = 4/5 for
part b). Thus the rate of change of speed of the collar at 0.75 m, is −9.81*0.6 = − 5.89 m/s2.
(1 mark for determining the sine of the angle, and hence at.)
b) The equation of motion in the normal direction is:
∑F
n
= N − mg cosθ = man = mv 2f / ρ so that: N = m(v 2f / ρ + g cosθ )
(1 mark)
3/2
!"1+ x 2 #$
Radius of curvature at x = 0.75 m, ρ (x) =
1
= 1.953 m @ x = 0.75m . (1 mark)
From the given geometry, ds = dy/sinθ, and so:
∫
vf
v0
v dv =
∫
at ds =
∫
at dy / sin θ = − ∫
y(0.75)
0
g dy
(using the result from a)
i.e.
v f = v02 − 2gΔy ≈ 2.987 m/s
(2 marks)
Thus: N = 2(2.9872 /1.953+ 9.81⋅ 0.8)≈ 24.8 N (1 mark for the numerical answer)