# Inertia Moment (Homework 3-103)

1. A hollow sphere (M = 3 kg, R = 60 cm) is rolling down an inclined surface with a height of 2.0 meters and an inclined angle of 25 degrees.a) find the moment of inertia of the sphere.

b) find the speed of the center of the mass when it reaches the bottom of the hill.

c) find the rotational kinetic energy at the bottom

d) find the linear acceleration of the sphere

e) for the sphere to make pure rotation find the magnitude of the static frictional coefficient.

2. A hollow spherical ball is let go from the top of a track and goes around a circular loop (R = 60 cm).

a) What must be the minimum height of the track so that the ball will not fall off the track at the topmost point of the loop?

b) when the above condition is met, what is the tangential speed of the ball when it reaches the topmost point?

3. A bucket (m = 4 kg) is falling down a 25 meter deep well as shown in the figure. The cylindrical drum on which is winded the bucket rope has a mass of 28 kg and radius of 15 cm.

a) find the tension in the rope.

b) find the linear acceleration of the descending bucket.

c) find the angular acceleration of the drum

d) find the angular velocity of the cylinder when the bucket touches the bottom of the well.

1.

a) The moment of inertia of a hollow sphere of mass M and radius R is

$I_sphere = 2*M*R^2/3 = 2*3*0.6^2/3 = 0.72 kg*m^2$

b. The total energy of the sphere at up and at the bottom is the same. Up there is only potential energy, at the bottom there is kinetic energy of rotation +kinetic energy of the center of mass

$M*g*H = I*\omega^2/2 +M*V(cm)^2/2$, $\omega$ is the angular speed of the sphere

the contact point of the spere with the plane has instant speed zero, the diametrical opposite point has instant speed $v = \omega*R,$ hence the instant speed of the center of mass is

$V(cm) = v/2 =omega*R/2$

$\omega = 2*V(cm)/R$

$M*g*H = I*4*V(cm)^2/R^2/2 + M*V(cm)^2/2$

$M*g*H = [I*4/R^2/2+M/2]*V(cm)^2$

$M*g*H = [(2M/3)*2 +M/2]*V(cm)^2$

$9.81*2 =[(2/3)*2 +1/2]*v(cm)^2$

$19.62 =1.83*V(cm)^2$

$V(cm) =3.27 m/s$

c. Rotational kinetic energy is

$E = I*\omega^2/2 = I*4*V(cm)^2/R^2/2 =(2M/3)*2*V(cm)^2 =(2*3/3)*2*3.27^2 =42.77 J$

4. V(cm)^2 =2*a*d, where d is the sliding distance down the inclined plane

$d = H/sin(\alpha) =2/sin(25) =4.73 m$

$a = V(cm)^2/2/d =3.27^2/2/4.73 =1.13 m/s^2$

d. The total friction force is equal to the friction force

$M*a = \mu*M*g$

$\mu = a/g = 1.13/9.81 =0.115$

2.

The condition of not falling is

$m*v^2/R = m*g$ (centrifugal force is equal to the weight)

$V^2/R = g$, or $V =\sqrt{(R*g)} = \sqrt{(0.6*9.81)} = 2.42 m/s$

The energy at the top of the loop (kinetic + potential) is equal to the energy at the maximum height (only potential)

$m*v^2/2 +m*g*(2R) = m*g*H$

$H =(V^2/2/g + 2R) =2.42^2/2/9.81 +2*0.6 = 1.498 m$

3.

considering the drum a solid cylinder it has a moment of inertia

$I = m*R^2/2 =28*0.15^2/2 =0.315 kg*m^2$

for the bucket one can write

$F +m*a = m*g$ (F is the force in the string)

For the drum one can write

$F = I*\epsilon$ (epsion is the angular acceleration of the drum)

$a =\epsilon*R$

Hence

$F = m*(g-a)$

$F= I*a/R$

or $I*a/R =mg-ma$

$(I/R+m)a =m*g$

$(0.315/0.15 +4)*a =4*9.81$

$a =6.43 m/s^2$

the force in the rope is

$F =m(g-a)= 4*(9.81-6.43) =13.52 N$

angular acceleration $\epsilon = a/R =6.43/0.15 =42.86 1/s^2$

tangential velocity of the cylinder is

$v= \sqrt{(2*a*s)} =\sqrt{(2*6.43*25)} = 17.93 m/s$

angular velocity is

$\omega =v/R =17.93/0.15 =119.54 1/s^2$

Bassim Elsamaloty

April 22, 2017 @ 7:56 pm

I believe there is a mistake in the formula for 2a. It does not appear that the rotational kinetic energy is considered in the conservation of energy formula.

I got $E1$ = $E2$

$E1$ = Potential energy at the top of the hill

$E2$ = The kinetic energy at the top of the loop + potential at the top of the loop + the rotational energy of the ball itself

Therefore:

$mgh = 1/2mv^2 + mg(2r) + (1/2)I\omega^2$

Making appropriate substitutions and simplification:

$h = 5/2r + \beta/2r$

Therefore

$h = 1.498 + 0.2 = 1.7$ Meters

valentin68

April 22, 2017 @ 8:11 pm

Thank you for your observation. I am just a human and making mistakes sometimes. Yes I forgot to consider the rotational energy.

Matt

April 21, 2018 @ 11:32 pm

You have quite a few of these wrong by the way. You are forgetting on the first equation that there is a 25 degree angle which means that the normal force is equal to Fn = mgcos(theta). For the frictional coefficient, you can’t simply just divide a/g. It has to be 2/3a/gcos25. A few others on here are wrong as well.