Question

A hoop rolls down a 4.75 m high hill without slipping. What is the final speed of the hoop, in meters per second?

Answer 1

Answer:6.82 m/s

Explanation:

Given

Height of hill $h=4.75\ m$

Suppose r is the radius of hoop and m be its mass

So moment  of inertia is given by

$I=mr^2$

Using conservation of energy we get

Potential energy at top=Kinetic+Rotational energy at bottom

$mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega ^2$

where $\omega =\frac{v}{r}$

v=velocity of hoop

$2gh=v^2+r^2\times \frac{v^2}{r^2}$   (without slipping)

$v^2=gh$

$v=\sqrt{gh}$

$v=\sqrt{98\times 4.75}$

$v=6.822\ m/s$

Answer 2

The final speed of the hoop, in meters per second is 9.67 m/s

Speed and distance based problem:

What information do we have?

Height of hill = 4.75

Using Third equation of motion

v² = u² + 2gs

Where,

v = final speed

u = initial speed

g = gravitational force

s = Disatnce

v² = (0)² + 2(9.8)(4.75)

v² = 0 + 2(9.8)(4.75)

v² = 93.5

final speed = 9.67 m/s

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