A 1.27 kg hoop with a radius of 10.0 cm rolls without slipping and has a linear speed of 1.32 m/s.
Find the rotational kinetic energy.
Find the total kinetic energy of the hoop.
First, you need the moment of interia for a hoop. Look it up.
rotationalKE=1/2 I w^2=1/2 I (v/r)^2
total KE=rotationalKE+1/2 m v^2
To find the rotational kinetic energy of the hoop, you can use the formula:
Rotational kinetic energy (KE_rot) = (1/2) * moment of inertia * angular velocity^2
The moment of inertia (I) of a hoop about its axis of rotation is given by:
I = m * r^2
where "m" is the mass of the hoop and "r" is the radius of the hoop.
Given:
Mass of the hoop (m) = 1.27 kg
Radius of the hoop (r) = 10.0 cm = 0.1 m
Linear speed of the hoop (v) = 1.32 m/s
First, let's calculate the moment of inertia:
I = m * r^2
I = 1.27 kg * (0.1 m)^2
I = 0.0127 kg * m^2
Next, we need to find the angular velocity (ω) of the hoop. Since the hoop rolls without slipping, the linear speed of the hoop is related to the angular velocity by:
v = ω * r
Rearranging the equation, we get:
ω = v / r
Substituting the values, we get:
ω = 1.32 m/s / 0.1 m
ω = 13.2 rad/s
Now, we can calculate the rotational kinetic energy:
KE_rot = (1/2) * I * ω^2
KE_rot = (1/2) * 0.0127 kg * m^2 * (13.2 rad/s)^2
KE_rot = 1.067 kg * m^2/s^2
Therefore, the rotational kinetic energy of the hoop is approximately 1.067 Joules.
To find the total kinetic energy of the hoop, you need to consider both its translational kinetic energy and rotational kinetic energy.
The translational kinetic energy (KE_trans) of the hoop is given by:
KE_trans = (1/2) * m * v^2
Substituting the given values, we get:
KE_trans = (1/2) * 1.27 kg * (1.32 m/s)^2
KE_trans = 1.756 kg * m^2/s^2
To find the total kinetic energy (KE_total), we need to add the translational kinetic energy and the rotational kinetic energy:
KE_total = KE_trans + KE_rot
KE_total = 1.756 Joules + 1.067 Joules
KE_total = 2.823 Joules
Therefore, the total kinetic energy of the hoop is approximately 2.823 Joules.