Question

Solve for the final angular velocity (ans. 22.9 rad/s)

Let's look at a new example to see how this works. Consider a solid wheel of radius 0.5 m and mass of 2 kg. The wheel sits at the top of a hill that has a height of 10 m. Starting from rest, the wheel rolls without slipping down the hill to the bottom. What is the velocity of the wheel at the bottom of the hill? The key to this problem is the statement "rolls without slipping". This is telling us that the velocity of the center of mass of the object is related to the angular velocity. Figure 1 Figure 1 shows a wheel rolling to the right. The red mark denotes a specific point on the wheel. After the wheel has made one complete rotation, the center of the wheel will have moved to the right by a distance equal to the circumference of the wheel. As the wheel rotates by some angle, the center of the wheel covers a distance equal to the are length

Answer 1

h Jom In the given problem wheel rolls without sliping then hill can be considered as inclined plane having height, tom. : the motion is mile without sliping so centre of most of whool moves in a straight line ) allume As the to whool stan to row Loth it having angular klocity W' and linear relocity 'le and moment of indeliai and a sadius of gysetion to them Accokeling the Conservation of every mgn I mis² + 1 W2 where fa mk2 and Wall- Ingin Ime - (114) momenty Pnecting of a Rinplwheel is given all fomitucentre) = Mp? mi? solkar) Henge – 9h = tg (2) + 9 = 1 What g = 9.8 m/s2 aw halom 2 Scanned with CamScanner...

NALANDA") Onpleting Abelles o the J 9.8x10 = 198= 9.899 mis Now Wawr som W le . .899 0.5 19.79 Jeals . . So angulen Kellty of wchall at holomak 19.79 bls.. Scanned with CamScanner
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