Dynamics Fall semester 2015Midterm exam 15.10.2015- type A

g

v

u

m(t)

1.1 (2 points) A rocket of initial mass mi is launchedfrom rest in the vertical direction. It burns fuel massaccording to the law m(t) = mi − αt, where α > 0.The gasses are expelled with a downward constant veloc-ity of magnitude u = |u|. Which of the following rela-tions represents the condition to have lift-off at time t =0?

(a) α > 0

(b) α >mig

u

(c) α <mig

u

(d) α > mig

k

mg

1.2 (1 point) A spring of stiffness k is com-pressed by ∆s to launch a particle of mass m verti-cally. What is the height hmax reached by the par-ticle?

(a) hmax =1

2

k∆s2

mg

(b) hmax = 2∆s

(c) hmax =2k∆s2

mg

(d) hmax =k∆s2

mg

1-1

1-2

g

h

1.3 (1 point) A particle of mass m is connectedto an unstretched spring at the initial height h. Forwhat value of the spring constant k does the massreach the ground with a velocity v?

(a) k =[mgh− 0.5mv2]

h2

(b) k =[mgh− 0.5mv2]

2h2

(c) k =[mgh− 0.5mv2]

4h2

(d) k =2 [mgh− 0.5mv2]

h2

v0

µ

↵

m

gx

h

2L

L

1.4 (2 points) A particle of mass m isreleased from a height h down an inclinedplane with an inclination angle α. At thebottom of the plane, there is a spring ofunstretched length 2L. The inclined planeis frictionless, except for its last section oflength L, where the friction coefficient isµ. What is the maximum downward initialvelocity v0 of the particle for its motion tobe conservative?

(a) v0 ≤

√1

m

[1

2kL2 −mgh− µmgL

]

(b) v0 ≤

√2

m

[1

2kL2 −mgh

]

(c) v0 ≤

√1

m

[1

2kL2 −mgh

]

(d) v0 ≤

√1

m

[1

2kL2 −mgh+ µmgL

]

1-3

x1

x2y2

y1

my

(t)

1.5 (2 points) A particle of mass m is at rest inthe horizontal x1 − x2 inertial frame. An observeris fixed to the rotating y1− y2 frame whose angularvelocity is θ = cos(t). No active force is applied tothe particle. Which of the following statements iscorrect?

(a) The observer in the y1 − y2 frame observes azero Coriolis force on the particle.

(b) The observer in the y1 − y2 frame observes a nonzero active force on the particle.

(c) The observer in the y1 − y2 frame observes a nonzero Euler force on the particle.

(d) None of the other statements is correct.

x1

x2y2

y1

my

(t)

1.6 (2 points) A particle of mass m is con-strained to move in a frictionless pipe attached toa frame rotating with constant angular velocity Ω,such that θ(t) = Ωt. No gravity is acting on the sys-tem. Which of the following statements is correct?

(a) The pipe exerts a constraint force on the par-ticle that balances the Coriolis force.

(b) The pipe exerts a constraint force on the par-ticle that generates the Coriolis force.

(c) The pipe exerts a constraint force on the particle that balances the Euler force.

(d) The pipe is smooth and hence exerts no constraint force on the particle.

1-4

g

h

m2m

A B

µ

↵

1.7 (2 points) Two blocks A and B of massmA = m and mB = 2m, respectively, are releasedfrom rest from the same height down a rough, in-clined plane with friction coefficient µ and inclina-tion angle α = π/4. Consider a dynamic Coulombfriction model, i.e. Ff = µN , where Ff is the mag-nitude of the friction force, and N is the magni-tude of the normal force between the sliding sur-faces. Which of the following statements is correct?

(a) Block A will reach the bottom before block B.

(b) Block B will reach the bottom before block A.

(c) Block A and block B will reach the bottom at the same time.

(d) Either (a) or (b) holds, depending on the value of h.

x

y

m

g

F

L

k O

1.8 (3 points) A particle of mass m is sus-pended from point O with a massless bar of lengthL and swings on a vertical plane. A force F of con-stant magnitude and normal to the bar is applied tothe mass. A torsional spring of stiffness k providesa restoring moment M = −kθ. Which is the correctequation of motion of the system?

(a) mL2θ + kθ − FL+mgL cos θ = 0

(b) mL2θ − kθ − FL−mgL sin θ = 0

(c) mL2θ + kθ − FL+mgL sin θ = 0

(d) mL2θ − kθ − FL−mgL sin θ = 0

1-5

m

gv0

R

/4

A

B

1.9 (3 points) A particle of mass m is releasedwith a velocity v0 from point A on the inclined,frictionless track sketched in the figure. Which isthe magnitude v0 of the initial velocity that makesthe particle leave the track at point B?

(a) v0 =√

5gR

(b) v0 =√

2gR

(c) v0 =√gR

(d) v0 =√

3gR

(e) v0 = 3gR

1.10 (2 points) A bullet is fired in Zurich to hit a target placed 500 meters to theEast. What is the expected hit location relative to the the center of the target?

(a) (b) (c) (d) (e)

1-6

x

y

v0

m

k

v1

2L0

L0

1.11 (3 points) A particle of mass m, slidingon a frictionless horizontal plane, is attached via aspring of stiffness k and unstretched length L0 to afixed point O. At time t = 0, the mass has a veloc-ity v0 perpendicular to the spring, and the springlength is 2L0. What is the velocity component ofthe mass along the spring direction v1‖ when thespring becomes unstretched?

(a) v1‖ =

√k

mL20 − v20

(b) v1‖ =

√k

mL20 + v20

(c) v1‖ =

√k

mL20 − 3v20

(d) v1‖ =

√k

mL20 − 3v0

x

y

m

g

L

O

R

1.12 (2 points) The pendulum shown in the fig-ure swings on a vertical plane. Denote by R themagnitude of the constraint force R. Which of thefollowing statements is correct?

(a) R = mg cos θ always.

(b) R > mg cos θ always.

(c) R < mg cos θ always .

(d) R depends on the motion θ(t).

1-7

1.13 (3 points) A spring of constant k is un-stretched when the slider of mass m reaches po-sition B. If the slider is released from restin position A, determine the normal force Nexerted by the guide on the slider at posi-tion C. Neglect friction between the mass andthe circular guide, which lies in a verticalplane.

(a) N = m

[5g +

kR

m(3− 2

√2)

]

(b) N = m

[3g +

kR

m(3− 2

√2)

]

(c) N = m

[4g +

km

R(3− 2

√2)

]

(d) N = m

[3g +

kR

m(3−

√2)

]

1.14 (2 points) The ball is released from posi-tion A with a velocity v0 and swings in a verticalplane. At the bottom position, the cord strikes thefixed bar at B, and the ball continues to swing alongthe dashed arc. What is the velocity of the ball whenit reaches position C?

(a) vC =√g(h−Rcosθ) + 1

2v20

(b) vC =√

2g(h+R(1− cosθ) + v20

(c) vC =√

2g(h−R(1− cosθ) + v20

(d) vC =√

2g(h−Rcosθ) + v20

1-8

.