Answered

Find expert answers and community support for all your questions on IDNLearn.com. Join our Q&A platform to get accurate and thorough answers to all your pressing questions.

5. A uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring). (1) M Before Wi (II) wf V Ө ə Ө After Ө wf H ? (a) What kind of frictional force acts on the ring upon contact with the surface? (b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), find the coefficient of friction corresponding to the frictional force you mentioned in (a). (c) What is the increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly? The ring then rolls smoothly up a ramp of 0 = π/6 rad and H = 5 m [see figure (II)] (d) What is the horizontal distance, from the end of the ramp, at which the ring lands?

5 A Uniform Solid Ring Of Mass M 10 Kg And Radius R 2 M Is Initially Rotating In Midair With Angular Speed 20 Rads The Before State In Figure I Below It Is Then class=
5 A Uniform Solid Ring Of Mass M 10 Kg And Radius R 2 M Is Initially Rotating In Midair With Angular Speed 20 Rads The Before State In Figure I Below It Is Then class=

Sagot :

(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.

(b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), the coefficient of friction corresponding to the frictional force is 0.612

(c) The increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly is 7500 J.

(d) The horizontal distance, from the end of the ramp, at which the ring lands is 7.78 m

What is frictional force?

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.

Given is a uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring).

(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.

(b) angular frequency = velocity / radius

ωf = 10/2 = 5 rad/s and ωi = 20 rad/s

Angular acceleration, α = ωf - ωi /t

Put the values, we get

α = -15/5 = -3 rad/s²

Coefficient of friction, μ = a/g = rα/g

Plug the values, we get

μ = 0.612

Thus, the coefficient of friction corresponding to the frictional force is 0.612.

(c) The energy lost = heat generated

 energy lost = 1/2 Iω² + 1/2 Mv²

 energy lost = 1/2 MR²ω² + 1/2 Mv²

Plug the values, we get

 energy lost = 7500 J

Thus, the  increase in the thermal energy is 7500 J.

(d) The horizontal distance, from the end of the ramp, at which the ring lands

s= (v- 2gH) sin2θ /g

s = (10 - 2x (-9.81)x5 ) sin (2x π/6) / 9.81

s = 7.78m

Thus, the horizontal distance is  7.78 m

Learn more about friction force.

https://brainly.com/question/1714663

#SPJ1

We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Your search for solutions ends here at IDNLearn.com. Thank you for visiting, and come back soon for more helpful information.