A ball of radius R=57cm and mass M=0.46kg is compressed against a spring of spring constant k=400N/m by an amount x=31.3cm measured from the equilibrium of the spring as shown in the figure below. The system is released from rest and when the spring reaches its equilibrium position the ball loses contact with the spring. Assume the ball rolls without slipping along the theta=250 incline after it is released. Find the translational speed of the ball after it travels d= 62.2cm from the point of losing contact with the spring . Isphere =2/5 MR2. Take g=9.8 m/s2 and express your answer using two decimal places.
To find the translational speed of the ball after it travels a certain distance, we can use conservation of energy. Let's start by finding the potential energy stored in the compressed spring.
The potential energy stored in a spring is given by the formula:
PE = (1/2)kx^2
where PE represents potential energy, k is the spring constant, and x is the compression (distance from the equilibrium position).
Substituting the given values into the formula:
PE = (1/2)(400 N/m)(0.313 m)^2
PE = 7.839 J
Next, let's calculate the potential energy at the point where the ball loses contact with the spring. This potential energy is then converted into kinetic energy as the ball rolls down the incline.
The potential energy at the point of losing contact is equal to the potential energy at the compressed position:
PE = 7.839 J
The total mechanical energy (E) at that point is the sum of the potential energy and the kinetic energy:
E = PE + KE
Since the ball rolls without slipping, the kinetic energy consists of both translational and rotational kinetic energy:
KE = (1/2)Mv^2 + (1/2)Iω^2
where M represents mass, v represents translational velocity, I represents moment of inertia, and ω represents angular velocity.
The moment of inertia for a solid sphere rolling without slipping is given by the formula:
I = (2/5)MR^2
where R represents the radius of the sphere.
Rearranging the equation for the moment of inertia:
ω = v/R
Substituting this equation into the kinetic energy formula:
KE = (1/2)Mv^2 + (1/2)(2/5)MR^2(v/R)^2
KE = (1/2)Mv^2 + (1/5)Mv^2
KE = (7/10)Mv^2
Now, equating the total mechanical energy with the sum of potential and kinetic energy:
E = PE + KE
Inserting the known values:
7.839 J = 7.839 J + (7/10)Mv^2
Simplifying the equation:
0 = (7/10)Mv^2
Since the mass and acceleration due to gravity are positive values, we can disregard the trivial solution v = 0.
Therefore, v = 0 cm/s.
The translational speed of the ball after it travels 62.2 cm from the point of losing contact with the spring is 0 cm/s.