1 Answer

Tin Bum · Answer 1 · 2022-04-14T11:12:38+0000

Answer:

d= 23.25 m

Step-by-step 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)

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:

$K_(rot) = (1)/(2)* I * \omega^(2) (6)$

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:

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

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

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

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