What linear speed must a 0.0495-kg hula hoop have if its total kinetic energy is to be 0.113 J? Assume the hoop rolls on the ground without slipping.

The kinetic energy of the hoop consists of two parts: (1) translational motion of the center of mass, (1/2) M V^2, and

(2) rotation about the center of mass,
(1/2)I w^2 = (1/2)M R^2*(V/R)^2=(1/2)MV^2
The two parts are equal in this case. The total kinetic energy is M V^2.
Solve
(0.0495)V^2 = 0.113
to obtain V. The answer will be in m/s.

To find the linear speed of the hula hoop, we need to apply the concept of kinetic energy and rotational motion. The given information that the hoop rolls on the ground without slipping implies that both linear and rotational speeds are related to each other.

First, let's calculate the rotational kinetic energy (K_rot) of the hula hoop. The rotational kinetic energy is given by the formula:

K_rot = (1/2) * I * ω^2

Where:
K_rot = Rotational kinetic energy
I = Moment of inertia of the hula hoop
ω = Angular speed of the hula hoop

Since the hoop rolls without slipping, we can relate the angular speed ω to the linear speed v using the formula:

v = ω * r

Where:
v = Linear speed of the hula hoop
r = Radius of the hula hoop

Now, let's rearrange the formula for ω using the relationship between linear and angular speed:

ω = v / r

Substituting this value back into the formula for the rotational kinetic energy:

K_rot = (1/2) * I * (v / r)^2

Given that the total kinetic energy (K_total) is 0.113 J, we can equate the total kinetic energy to the sum of linear and rotational kinetic energies:

K_total = K_lin + K_rot

0.113 J = K_lin + K_rot

Since the hula hoop only has rotational kinetic energy when it rolls, the linear kinetic energy is equal to the total kinetic energy minus the rotational kinetic energy:

K_lin = K_total - K_rot

Now, we can write the equation for K_lin using the formula for linear kinetic energy (K_lin = (1/2) * m * v^2) and rearrange it to solve for v:

(1/2) * m * v^2 = K_total - ((1/2) * I * (v / r)^2)

Plugging in the given values:
m = 0.0495 kg (mass of the hula hoop)
K_total = 0.113 J (total kinetic energy of the hula hoop)
r = radius of the hula hoop

We can substitute these values into the equation and solve for v.