Angular Speed of Rolling Sphere: 32o Roof, L = 7.0 m, 12 kg, 13 cm

[melissa_y]

A solid sphere of radius 13.0 cm and mass 12.0 kg starts from rest and rolls without slipping a distance of L = 7.0 m down a house roof that is inclined at 32o. What is the angular speed about its center as it leaves the house roof? Use units of "rad/s".

 

[himanshu121]

the Forum helps u, if u show us that u have worked upon anyway
solve torque equation
$$\Sigma \tau_{cm} =I \alpha$$
$$\Sigma F= Ma$$
Use condition of rolling
$$a_{cm}=\alpha R$$

 

[thepatientmental]

To calculate the angular speed of the rolling sphere, we can use the formula ω = v/R, where ω is the angular speed, v is the linear speed, and R is the radius of the sphere.

First, we need to calculate the linear speed of the sphere as it rolls down the roof. We can use the formula v = √(2gh), where g is the acceleration due to gravity and h is the height of the roof. Plugging in the given values, we get v = √(2*9.8*7.0*sin(32)) = 8.11 m/s.

Next, we can calculate the radius of the sphere using the given diameter of 13 cm. Converting to meters, we get R = 0.13 m.

Finally, we can plug in these values into the formula ω = v/R to get the angular speed of the sphere as it leaves the roof:

ω = 8.11/0.13 = 62.38 rad/s

Therefore, the angular speed of the rolling sphere is 62.38 rad/s as it leaves the house roof.