Question

A lawn roller in the form of a solid uniform cylinder is being pulled horizontally by a horizontal force B applied to an axle through the center of the roller. The roller radius 0.59 m and mass 48 kg. What magnitude B of the force is required to produce and acceleration a=4.2 m/s^2 of the center of mass of the roller if the lawn rolls without slipping?

Answer 1

Answer:

Explanation:

mass, m = 48 kg

radius, r = 0.59 m

acceleration of center of mass, a = 4.2 m/s2

The free body diagram of cylinder is as follows:

Since the cylinder is rolling without slipping, therefore,a=αr,

which gives, angular acceleration, α = a/r = (4.2/0.59)rad/s2 = 7.12 rad/s2

Moment of inertia of cylinder about center of mass C, IC= (1/2)mr2

So, by parallel axis theorem, moment of inertia about point A,IA = (1/2)mr2 + mr2 =(3/2)mr2 = (3/2)*48*0.592 kg.m2 =25.06 kg.m2

Now, taking the moment of forces about point A, T =IAα

which gives, B*r = IAα, and we get, B =(25.06*7.12)/0.59 N = 302.42 N

Answer 2

The magnitude B of the force that is required to produce and acceleration a=4.2 m/s^2 is 302.42 N

Calculations and Parameters:

• mass, m = 48 kg
• radius, r = 0.59 m
• acceleration of center of mass, a = 4.2 m/s2

To find the angular acceleration, we would solve:

α = a/r = (4.2/0.59)rad/s2 = 7.12 rad/s2

We should also note that the moment of inertia of cylinder about center of mass C, IC= (1/2)mr2

Hence, we would solve:

A,IA = (1/2)mr2 + mr2 =(3/2)mr2 = (3/2)*48*0.592 kg.m2 =25.06 kg.m2

If we take these moment of forces at a point A is T =IAα

Finally, we would solve:

B*r = IAα,

B =(25.06*7.12)/0.59 N = 302.42 N

Read more about moment of inertia here:

https://brainly.com/question/15248039