Answer:
The acceleration of its center of mass is
The frictional force is [tex]f = 21.65 \ N[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m = 17 kg[/tex]
The radius of the ball is [tex]r = 50cm = \frac{50}{100} = 0.5m[/tex]
the angle with the horizontal is [tex]\theta = 19 ^ o[/tex]
The of the ball at the base is [tex]v = 5.0 \ m/s[/tex]
This setup is shown on the first uploaded image
looking at the diagram we see that the force acting on the ball can be mathematically evaluated as
[tex]mg sin \theta -f = ma[/tex]
Where f is the frictional force
The torque on the ball is mathematically represented as
[tex]\tau = f * r[/tex]
This torque can also be mathematically represented as
[tex]\tau = I \alpha[/tex]
where I is the moment of inertia of the ball which is mathematically represented as
[tex]I = \frac{2}{3} m r^2[/tex]
While [tex]\alpha[/tex] is the angular acceleration which is mathematically represented as
[tex]\alpha = \frac{a}{r}[/tex]
So [tex]\tau = \frac{2}{3} m r^2 * \frac{a}{r}[/tex]
Equating the both formula for torque
[tex]f * r = \frac{2}{3} m r^2 * \frac{a}{r }[/tex]
=> [tex]f = \frac{2}{3} ma[/tex]
Substituting this for f in the above equation
[tex]mg sin \theta = ma + \frac{2}{3} ma[/tex]
[tex]g sin \theta = \frac{5}{3} a[/tex]
[tex]a = \frac{3}{5} * g * sin \theta \alpha[/tex]
Substituting values
[tex]a = 1.91 m/s^2[/tex]
Now substituting into the equation frictional force equation
[tex]f = \frac{2}{3} * 17 * 1.91[/tex]
[tex]f = 21.65 \ N[/tex]
Answer:
[tex]a=-1.92 m/s^{2}[/tex]
[tex]F_{f}=-21.76 N[/tex]
Explanation:
We can use the definition of the torque:
[tex]\tau=I\alpha[/tex]
When:
Now, torque is the product of the friction force times the radius.
[tex]F_{f}*R=\frac{2}{3}mR^{2}*\frac{a}{R}[/tex]
[tex]F_{f}=\frac{2}{3}ma[/tex] (1)
Now, let's analyze the force acting over the sphere using the Newton's second law.
[tex]F=ma[/tex]
[tex]-mgsin(\theta)-F_{f}=ma[/tex] (2)
Let's put F(f) of the equation (1) into the equation (2):
[tex]-mgsin(\theta)-\frac{2}{3}ma=ma[/tex]
[tex]a=-\frac{3}{5}gsin(\theta)[/tex]
[tex]a=-\frac{3}{5}*9.81*sin(19)[/tex]
[tex]a=-1.92 m/s^{2}[/tex]
Hence: [tex]F_{f}=\frac{2}{3}ma=\frac{2}{3}*17*(-1.92)[/tex]
[tex]F_{f}=-21.76 N[/tex]
I hope it helps you!
