Consider a wheel of radius 1:15 m, mass 8:7 kg and moment of inertia?

Question

Hint: Consider the wheel's energy. Consider a wheel of radius 1:15 m, mass 8:7 kg and moment of inertia I = (MR^2)/2 (it's a solid disk). The wheel rolls without slipping in a straight line in an uphill direction 17 degree above the horizontal. The wheel
starts at angular speed 18:3 rad/s but the rotation slows down as the wheel rolls uphill, and eventually the wheel comes to a stop and rolls back downhill. The acceleration of gravity is 9:8 m/s^2. How far does the wheel roll in the uphill direction before it stops? Answer in units of m

Answer 1

Mechanical energy is conserved. At the bottom of the hill the kinetic energy is 1/2 m v^2 + 1/2 I w^2 where m is mass, v velocity of the center of mass, I is moment of inertia, and w is angular velocity. Since the wheel rolls without slipping, v = r w. Potential energy is zero at this point.

Now the wheel goes up the hill. It will stop when all of the kinetic energy has been converted to potential: that is, when

mgh = 1/2 m v^2 + 1/2 I w^2

Solve for h. But you don't want height, you want distance traveled. This is found by

d sin 17 = h.

Answer 2

____________________________________-
Mass of wheel=m=8.7 kg

radius of wheel=r=1.15 m

The wheel is a disc.

Moment inertia of wheel (disc)=I=(1/2)mr^2=0.5*8/7*(1.15)^2

Moment inertia of wheel (disc)= I = 5.7529 kgm^2

The initial angular speed of wheel =wi =18.3 rad /s

The initial linear speed of wheel = v =rw =1.15*18.3 = 21.045 m/s

Kinetic energy of translational motion= KEt =(1/2)mv^2

Kinetic energy of translational motion= KEt =(1/2)*8.7*(21.045)^2=1926.59 J

Kinetic energy of rotational motion= KEr =(1/2)Iw^2=0.5*5.7529*(18.3)^2=963.29

Initial mechanical energy=1/2 m v^2 + 1/2 I w^2

Initial mechanical energy=1926.59 + 963.20 =2889.88 J

Total mechanical energy when wheel stops and is about to roll back=mgh

If 'h' is the vertical height and s is length of slant straight path; then,

h=s sin 17

Total mechanical energy at height=mgs sin 17=8.7*9.8* s*sin17=24.9276* s J

As mechanial energy is conserved,

mechanical energy at top = mechanical energy at bottom

24.9276 s= 2889.88

s =2889.88 /24.9275=115.9314 m

the wheel rolls 115.9314 m in the uphill direction before it stops
______________________________________________

Answer 3

2nd of inertia( I) is given by applying : I = (MR^2)/4 for a disc I = (MR^2)/2 for a hoop/hoop Subtituing the values you will get that Hoop has a much better 2nd of inertia(15.8) in assessment to disk with I = 15 !!