A cylindrical object of mass M and radius R rolls smoothly from rest down a ramp and onto a horizontal section. From there it rolls off the ramp and onto the floor, landing a horizontal distance d = 0.506 m from the end of the ramp. The initial height of the object is H = 0.90 m; the end of the ramp is at height h = 0.10 m. The object consists of an outer cylindrical shell (of a certain uniform density) that is glued to a central cylinder (of a different uniform density). The rotational inertia of the object can be expressed in the general form I= .

The question is not complete and the complete question is ;

A cylindrical object of mass M and radius R rolls smoothly from rest down a ramp and onto a horizontal section. From there it rolls off the ramp and onto the floor, landing a horizontal distance d = 0.506 m from the end of the ramp. The initial height of the object is H = 0.90 m; the end of the ramp is at height h = 0.10 m. The object consists of an outer cylindrical shell (of a certain uniform density) that is glued to a central cylinder (of a different uniform density). The rotational inertia of the object can be expressed in the general form I = BMR², but B is not 0.5 as it is for a cylinder of uniform density. Determine B.

β = 1.36

Explanation:

First of all, the speed of the object at the bottom of the ramp is the same as its speed when it is launched into the air. This can be found by working a projectile motion. So, we'll choose a coordinate system with the origin on the floor directly under the end of the ramp, in the positive x direction where the object is launched, and positive y-direction upward.

For y-direction, time of flight is gotten from;

yf = yi + viy(t) + ½(ay) t²

where;

yf = 0 m

yi = h = 0.10 m

viy = 0 m/s

ay = -g = -9.8 m/s²

Substituting the values, we obtain ;

0 = 0.1 + 0 + ½(-9.8) t²

4.9t² = 0.1

t = √(0.1/4.9)

t = 0.1429 s

Now, for the x direction

xf = xi + vix t + ½(ax) t²

Where;

xf = 0.506 m

xi = 0m

ax = 0 m/s²

t = 0.1429 s

Substituting the relevant values and solving for vix, we have;

vix = xf /t = (0.506 m) / (0.1429 s) = 3.54 m/s

Thus, the object's center of mass is traveling at 3.54 m/s when it reaches the bottom of the ramp.

At this point, vf will now be 3.54 m/s since the initial speed when the object is launched is the final speed when it rolls down the ramp.

Using the Work-Energy Theorem to analyze the motion of the object rolling down the ramp,we get :

Wext + Wnc = ΔK + ΔU

where;

Wext = the work done by external forces on my system,

Wnc is the work done by non-conservative forces

ΔK is the change in kinetic energy ΔU is the change in potential energy.

In this system of the object, earth and ramp, Wext and Wnc are zero. Thus, the only potential energy is that due to the gravitational force between the Earth and the object while change in the Earth's kinetic energy is negligible.

Therefore, ΔK can be taken to be entirely due to the object.

Now, ΔK will be represented as a translational kinetic energy ½mv² and a rotational kinetic energy ½Iω² about the center of mass.

Thus; from Wext + Wnc = ΔK + ΔU;

0 + 0 = [ ½m(vf)² - ½m(vi)² ] + [ ½I(ωf)² - ½I(ωi)² ] + [mgh(f) - mgh(i)]

Since the object starts at rest, vi and ωi are zero.

Also, let's subtitute the following ;

hf = h ; hi = H ; m = M

0 = [ ½M(vf)² - 0 ] + [ ½I(ω_f)² - 0 ] + [ Mgh - MgH ]

0 = [ ½M(vf)²] + [ ½I(ωf)²] + [ Mgh - MgH]

Now, since the object rolls without slipping about its outer circumference, angular frequency ωf = vf/ R.

From the question, I = βMR², so

0 = ½M(vf)² + ½(βMR²)(vf/R)² + [Mgh - MgH]

0 = ½M(vf)² + ½(βM)(vf)² + [Mgh - MgH]

Divide each term by M to obtain;

0 = ½(vf)² + ½(β)(vf)² + [gh - gH]

Multiply each term by 2 to obtain ;

0 = (vf)² + (β)(vf)² + 2[gh - gH]

0 = (vf)² + (β)(vf)² + 2g[h - H]

0 = (1 + β)(vf)² + 2g [h - H]

We are looking for β so let's try to make it the subject;

(1 + β) = 2g[H - h]/(vf)²

β = 2g[H - h]/(vf)² - 1

β = 2(9.8)[0.9 - 0.1] / (3.54)² - 1 =1.36

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