The string constrains the rotational and translational motion of the falling cylinder, given that it doesn't slip. What is the relationship between the magnitude of the angular velocity ω and that of the velocity v of the center of mass of the cylinder? Express ω in terms of v and r.

ω = v / r

Explanation:

Since the cylinder is wrapped about the string, it rolls without slipping with respect to the string. The constraint  relationship to assure this kind of motion can be obtained by considering the relative velocity of the point on the  cylinder that is in contact with the string with respect to the string­­it should be zero.

The velocity of the center of the cylinder,

v = dy (t) / dt,  r and the  rotation rate of the cylinder, ω

We have to first find the velocity of the contact point relative to the center of the cylinder. Since this ignores the  translational motion of the entire cylinder with respect to the ground, we then need to add back in the cylinder's center­ of ­mass velocity to find the velocity of the contact point with respect to the ground.

So,

v (contact) = ωr - v

where,

ω is the angular velocity

v is the linear velocity

r is the radius

ω = v / r

the answer is a)

mark brainliest

the maxium height is 34m/s