Question

Consider a wheel (solid disk) of radius 1.12 m, mass 10 kg and moment of inertia 1 2 M R2 . The wheel rolls without slipping in a straight line in an uphill direction 37◦ above the horizontal. The wheel starts at angular speed 12.0536 rad/s but the rotation slows down as the wheel rolls uphill, and eventually the wheel comes to a stop and rolls back downhill. How far does the wheel roll in the uphill direction before it stops?

Answer 1

Answer:

d= 23.25 m

Explanation:

• Assuming no other external forces acting on the disk, total mechanical energy must be conserved.
• Taking the initial height of the disk as the zero reference for the gravitational potential energy, initially. all the energy is kinetic.
• This kinetic energy is part translational kinetic energy, and part rotational kinetic energy, as follows:

       $E_{o} = K_{transo} + K_{roto} (1)$

• When the disk rolling uphill finally comes to an stop, its energy is completely gravitational potential energy, as follows:

       $E_{f} = m*g*h (2)$

• Since the angle with the horizontal of the track on which the disk is rolling, is 37º, we can express the height h in terms of the distance traveled d and the angle of 37º, as follows:

       $h = d* sin 37 (3)$

• Replacing (3) in (2):

       $E_{f} = m*g* d * sin 37 (4)$

• Since the wheel rolls without sleeping, this means that at any time there is a fixed relationship in the translational speed and the angular speed, as follows:

       $v = \omega * R (5)$

• For a solid disk, as mentioned in the question, the moment of inertia is just 1/2*M*R².
• The rotational kinetic energy of a rotating rigid body can be written as follows:

       $K_{rot} = \frac{1}{2}* I * \omega^{2} (6)$

• Replacing I from (6) and ω from (5), and remembering the definition of the translational kinetic energy, we can solve (1) in terms of v, m and r as follows:

       $E_{o} = K_{transo} + K_{roto} = \frac{1}{2}* m* v^{2} +(\frac{1}{2}* \frac{1}{2}) *m*r^{2}*(\frac{v}{r}) ^{2} = \\ \frac{3}{4} * m * v^{2} (7)$

• Since (4) and (7) must be equal each other, we can solve for d as follows:

       $d =\frac{3}{4} * \frac{v^{2}}{g*sin37} = \frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} (8)$

• Replacing by the values, we finally get:

       $d =\frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} = \frac{3}{4} *\frac{(12.0536rad/sec*1.12m)^{2}}{9.8 m/s2*0.601} = 23. 25 m.$