November 7, 2007
PHY2053
Quiz 7 (Chapter 7)
Name:
UFID:
1. (5 pts) A wheel starts from rest and reaches an angular speed of 8.00 rad/s in 4.00 s.
Assume the angular acceleration is constant and determine the angle (in radians)
through which the wheel rotates during this period.
We calculate the angular acceleration of the wheel:
α = Δω/Δt = 8/4 = 2 rad/s²
Thus the angular displacement is
θ = (1/2)αΔt² = (1/2)×2×4² = 16 rad
2. (5 pts) A 0.50 kg ball that is tied to the end of a 2.00 m cord is revolved in a horizontal
plane so that its speed is 5.00 m/s. What angle does the cord make with the vertical?
Applying the Newton’s second law, we get
X-component: mv²/r = Tsinθ, Y-component: Tcosθ-mg = 0
Substitute r = Lsinθ into the first equation and solve both equation for T.
X: T = mv²/Lsin²θ = mv²/L(1-cos²θ) , Y: T = mg/cosθ
We equate both equation and solve it for cosθ:
mv²/L(1-cos²θ) = mg/cosθ ⇒ cos²θ +v²/(Lg)-1 = 0, where v²/(Lg) = 5²/(2×9.8) = 1.276
cosθ = (-1.276+√(1.276²+4))/2 = 0.548 (Since -1<cosθ<1, we ignore the other solution.)
Therefore, θ = cosˉ¹(0.548) = 56.8º
3. (5 pts) A 2.00 kg ball is released from the top of a hemispherical track with a speed of
4.00 m/s. The radius of the track is 5.00 m and you can neglect friction. At what position
does the ball leave the track? Express it as an angle measured from the vertical.
When the ball leaves the track, the normal force becomes zero. Applying Newton’s
second law, we get
m(V²/r) = mgcosθ ⇒ V² = rgcosθ
Since there is no work done by non-conservative forces, the mechanical energy is
conserved.
(1/2)mVi²+ mgr = (1/2)mV²+mgrcosθ ⇒ V² = Vi²+2gr(1-cosθ)
Equating these two equations and solving it for cosθ, we get
rgcosθ = Vi²+2gr(1-cosθ) ⇒ cosθ = Vi²/(3gr)+2/3 = 4²/(3×9.8×5)+2/3 = 0.7755
Therefore, θ = cosˉ¹(0.755) = 39.1º
4. (5 pts) Objects with masses of 200 kg and 600 kg are separated by 1.00m. Another
object with mass of 80.0 kg is placed midway between them. Find the position where the
net force on the 80 kg object is zero and express it as a distance from the 600 kg object.
Let X be the distance from the 600 kg object, then 1-X is the distance from the 200 kg
object. Since the net force on the third mass is zero at X, we get
-GmM/X² + GmM’/(1-X)² = 0 ⇒ -M(1-X)² + M’X² =0 ⇒ (M’-M)X²+2MX-M = 0
Substitute the given values and simplify the equation:
-400X²+1200X-600 = 0 ⇒ 2X²-6X+3 = 0
Using the formula for quadratic equations, we get
X = (6-√(6²-4×2×3))/(2×2) = 0.634m
(You have to exclude the other solution because 0<X<1.)