Question

A 240 N sphere 0.20 m in radius rolls, without slipping 6.0 m downa

ramp that is inclined at 28 degrees with the horizontal. What isthe
angular speed of the sphere at the bottom of the hill if itstarts
from rest, in rad/s?

Answer 1

The angular speed of the sphere at the bottom of the incline can be calculated using the conservation of mechanical energy, considering both translational and rotational kinetic energies.

Step-by-step explanation:

To find the angular speed of the sphere at the bottom of the incline, we can use the principle of conservation of mechanical energy. As the sphere rolls down the incline without slipping, its potential energy at the top is converted into translational kinetic energy and rotational kinetic energy at the bottom.

Let's denote the mass of the sphere as m, the radius as r, the acceleration due to gravity as g, the height of the ramp as h, the angle of inclination as θ, and the angular speed at the bottom as ω.

Using the relation for a sphere rolling without slipping, KE_{rotational} = (1/2)Iω^2, where I is the moment of inertia for a solid sphere, I = (2/5)mr^2, and kinetic energy KE_{translational} = (1/2)mv^2 where v = rω. We can set the potential energy at the top, PE = mgh, equal to the sum of rotational and translational kinetic energies at the bottom.

Solving for ω, we have ω = √((10gh)/(7r^2)). To find h, we use the trigonometric relation h = sin(θ)d, where d is the distance the sphere rolled down the ramp. Plugging in values yields h = sin(28°)(6.0m).

Finally, calculating the angular speed, we find ω = √((10*9.8*sin(28°)*6.0m)/(7*0.20m^2)).

Answer 2

The angular speed of the sphere at the bottom of the hill is 31.39 rad/s.

Step-by-step explanation:

It is given that,

Weight of the sphere, W = 240 N

Radius of the sphere, r = 0.2 m

Angle with the horizontal,
`\theta=28^(\circ)`

We need to find the angular speed of the sphere at the bottom of the hill if it starts from rest.

As per the law of conservation of energy, the total energy at the top is equal to the energy at the bottom.

Gravitational energy = translational energy + rotational energy

So,

`mgh=(1)/(2)mv^2+(1)/(2)I\omega^2`

I is the moment of inertia of the sphere,
`I=(2)/(5)mr^2`

Also,
`v=r\omega`

h is the height of the ramp,
`h=l\ sin\theta`

`mgl\ sin\theta=(1)/(2)m(r\omega)^2+(1)/(2)I\omega^2`

On solving the above equation we get :

`\omega=\sqrt{(10gl\ sin\theta)/(7r^2)}`

`\omega=\sqrt{(10* 9.8* 6\ sin(28))/(7(0.2)^2)}`

`\omega=31.39\ rad/s`

So, the angular speed of the sphere at the bottom of the hill is 31.39 rad/s. Hence, this is the required solution.