Mechanics 6 Questions

1) An object of mass 5.0 kg revolves at 600 rpm in a horizontal circle of radius 1.5 m. Determine the a) period of the object’s motion in seconds

b) object’s speed

c) magnitude of the centripetal acceleration

d) magnitude of the centripetal force acting on the object.

2) A 2.0 kg object is released from rest from the top of a 5.0 m long frictionless incline. The angle of the incline is 30 degree above the horizontal. Calculate the objects

a)velocity at the bottom of the incline

b)rate of acceleration and

c)time required for the object to reach the bottom.

3) A sphere of mass 2.0 kg is released from rest and rolls without slipping down a 5.0 m long incline which is elevated by 0.50 m at one end. Calculate the

a) sphere’s tangential speed at the bottom and

b) time to travel from the top to the bottom.

4)Andy and Bob are carrying Chuck on a 5.0 m long uniform horizontal plank which weights 200 N. Chuck weighs 500 N and sits 1.0 m from Bob. Calculate the magnitude and direction of the

a)force that Bob exerts and

b)force that Andy exerts.

5) A wooden raft has a density of 600 kg/m^3. The raft is 5.0 m long,4.0 m wide and 3.0 m high and floats on water.The density of water is 1000 kg/m^3.Calculate the height of the log which is above the water line.

6) A 0.50 kg object attached to a vertical spring causes the spring to stretch 0.10 m. The spring is displaced 0.050 m from the equilibrium and released.

a) Determine the force constant of the spring

b) period of motion

c)Derive a formula for the object’s motion

d)Determine the object’s speed and acceleration when it is 0.025 m from equilibrium


a) The period T of the motion is by definition

$T = 1/F$, where F is the frequency

$T =1/600/60 =2.77*10^{-5} s$

b) the speed v is again by definition ($\omega$ is the angular speed)

$v = \omega*R =2*\pi*F*R =2*\pi*(600/60)*1.5 = 94.24 m/s$

c) centripetal acceleration is

$a = \omega^2*R = v^2/R =94.24^2/1.5 = 5921.76 m/s^2$

d) centripetal force

$F = m*a = 5*5921.76 =29608.81 N$


The height of the incline is

$H = L*sin(\alpha) = 5*sin(30) = 2.5 m$

a) at the top there is only potential energy, at the bottom there is only kinetic energy. the total energy is the same.

$m*g*H = m*v^2/2$

$v = \sqrt(2*g*H) =\sqrt(2*9.81*2.5) = 7 m/s$

b) the acceleration a comes from the equation (L is the length of the incline)

$v^2 =2*a*L$

$a = V^2/2/L =7*7/2/5 = 4.9 m/s^2$

c) the time t comes from the equation

$v = a*t$

$t = v/a =49/4.9 = 10 s$

3.  a) same considerations as at 2.a)

$m*g*H = m*V^2/2$

$V = \sqrt{(2*g*H)} = \sqrt{(2*0.5*9.81)} = 3.13 m/s$

b) the acceleration comes from the equation (L is the length of the incline)

$v^2 =2*a*L$

$a= V^2/2/L = 0.981 m/s^2$

the time comes from the equation

$V =a*t$

$t = V/a =3.13/0.981 =3.19 s$

4. The rotational momentum of the plank is zero with respect to Andy.

$G_{plank}*(L/2)+G_{chuck}*(L-1) – F_{bob}*L = 0$

$F_{bob} = (200*2.5 +500*4)/5 = 500 N$

5. The mas of the raft is ($\rho$ is the density)

$Mraft = \rho(raft)*H*L*w =600*5*4*3 = 36000 kg$

The mass of the liquid displaced by the submerged part of the raft should equal the total mass of the raft.

$\rho(water)*h*L*w = M_{raft}$

$h = M_{raft}/L/w/\rho(water) =36000/5/4/1000 = 1.8 m$ submerged

the height of the raft above the water line is

$H1 = H-h =3 – 1.8 =1.2 m$


a) the force constant is $m*g =k*x$ (x is the displacement)

$k = m*g/x =0.5*9.81/0.1 = 49.05 N/m$

b) period of motion is

$T = 2*\pi*\sqrt{(m/k)} = 2*\pi*\sqrt{(0.5/49.05)} =0.63 s$

c) amplitude of motion is $A =0.05 m$

initial phase of motion is $\phi = 0 degree = 0 rad$

angular speed of the motion $\omega = 2*\pi/T = 2*\pi/0.63 = 9.9 rad/s^2$

the displacement x as a function of time t (in seconds) is

$x = A*sin (\omega*t +\phi) = 0.05*sin(9.9*t + 0) m$


$0.025 =A*sin(\omega*t)$

$\omega*t = \arcsin (0.025/0.05) =0.523$

$t =0.523/9.9 = 0.0528 sec$

$v = d x/d t = A*\omega*cos(\omega*t) =0.05*9.9*cos(9.9*0.0528) = 0.428 m/s$

$a = d v/d t = -A*\omega^2*sin(\omega*t) = -\omega^2*x = -9.9^2*0.025 =2.45 m/s/s$