(#1.) A top starting from 40 rad/s takes 200 rad to stop. What is its rotational acceleration? (a constant)

(#2.) A stationary 100 Kg circular object (r= 0.20 m) is on a 35 degree incline. The coefficient of rolling friction between the block and the incline is 0.01. If it moves for 5 seconds, what is the angular displacement of the object at this time? (disregard torque)

To answer question #1, we need to calculate the rotational acceleration of the top. We know that the final angular velocity of the top is 0 rad/s (since it stops) and the initial angular velocity is 40 rad/s. We also know that the angle covered is 200 rad.

We can use the following equation to find the rotational acceleration:

Final angular velocity (ωf) = Initial angular velocity (ωi) + (Rotational acceleration (α) * Angle covered (θ))

Plugging in the given values, we have:

0 rad/s = 40 rad/s + (α * 200 rad)

Simplifying the equation, we get:

-40 rad/s = α * 200 rad

Dividing both sides by 200, we get:

-40 rad/s / 200 rad = α

Therefore, the rotational acceleration of the top is -0.2 rad/s^2.

Moving on to question #2, we need to find the angular displacement of the object on the incline. Since there is no torque mentioned in the problem, we can assume that the object is rolling without slipping. In this case, the relation between linear displacement (x) and angular displacement (θ) is given by:

θ = x / r

where r is the radius of the circular object.

We are given the linear displacement x as 5 seconds, and the radius of the object is 0.20 m. Plugging in these values, we get:

θ = 5 s / 0.20 m

Simplifying the equation, we get:

θ = 25 rad

Therefore, the angular displacement of the object at this time is 25 rad.