A penny rests on top of a bowling ball. You give the bowling ball a slight bump and the penny slides off the frictionless, spherical surface as shown below. At what angle θ does the penny leave the surface of the bowling ball? (Hint: When the penny leaves the surface, the normal force is zero.) Assume the friction between the penny and the bowling ball is negligible.

48 degrees

Well, let's see. The penny leaves the surface of the bowling ball when the normal force is zero. So, we can say the penny is trying to escape the ball, probably because it wants to roll with a different crowd or something.

Now, since the surface is frictionless, there is no force trying to keep the penny on the ball. How would you feel if you were constantly rolling around on a big ball with no friction? It's definitely a slippery slope, my friend.

So, let's get back to the question. We need to find the angle θ at which the penny leaves the surface. Well, think about it this way - if the normal force is zero, that means the gravitational force is the only force acting on the penny at that moment. And we all know that gravity works straight down, just like your mood whenever you lose a game of bowling.

So, the penny is going to leave the surface of the bowling ball when the gravitational force overpowers the centripetal force. And since the penny is sliding down, it means the centripetal force is directed towards the center of the ball.

Now, I could throw a bunch of equations at you, but let's keep it simple. The centripetal force is represented by the equation mv²/r, where m is the mass of the penny, v is its velocity, and r is the radius of the ball. And since gravity is trying to steal the show here, we know the gravitational force is mg, where g is the acceleration due to gravity.

So, when the penny leaves the surface, the gravitational force mg is equal to the centripetal force mv²/r. Now, we know that the velocity of the penny is related to the angle θ, because as the penny slides down, it's going to pick up some speed.

Unfortunately, my inner comedian can't calculate these equations for you. Luckily, if you plug in the values for the mass of the penny, acceleration due to gravity, and radius of the ball, you can solve for the angle θ. Just make sure to triple-check your calculations, because we don't want any boomerang bowling balls flying around, do we?

To determine the angle θ at which the penny leaves the surface of the bowling ball, we can analyze the forces acting on the penny at that moment.

1. At the moment the penny leaves the surface, the only force acting on it is its weight, pointing downward.

2. We can resolve this weight force into two components: one parallel to the surface (tangential to the surface) and one perpendicular to the surface (normal to the surface).

3. The perpendicular component of the weight is balanced by the normal force exerted by the bowling ball, which is zero when the penny leaves the surface.

4. Therefore, the only force acting on the penny that can cause it to leave the surface is the tangential component of the weight.

5. The tangential component of the weight points along the direction of motion of the penny as it slides off the surface, tangential to a circle with the same radius as the bowling ball.

6. By using basic trigonometry, we can determine that the angle θ at which the penny leaves the surface is equal to the angle between the direction of motion (tangential to the circle) and the downward vertical direction.

Therefore, the angle at which the penny leaves the surface of the bowling ball is θ = 90 degrees or π/2 radians.

To determine the angle θ at which the penny leaves the surface of the bowling ball, we need to consider the forces acting on the penny at that moment. Since the normal force is zero when the penny leaves the surface, the only force acting on the penny is its weight (mg).

To find the angle θ, we can use trigonometry. Let's denote the angle between the vertical direction and a line connecting the center of the bowling ball to the point where the penny slides off as α.

Using trigonometry, we can relate angle θ and angle α:

tan(θ) = tan(α)

Now, we need to find α. Since the penny slides off the frictionless surface of the bowling ball, the path it follows after sliding off is tangential to the surface at that point. This means that the line connecting the center of the bowling ball to the point where the penny slides off is perpendicular to the tangent line at that point.

Therefore, α is the angle formed between the vertical direction and the perpendicular line to the tangent at the point where the penny slides off.

To find α, we need to find the tangent line at that point. Since the surface of the bowling ball is spherical, the tangent line at any point on its surface is perpendicular to the radius at that point.

When the penny slides off, it leaves the surface along a radius. Hence, the tangent line at that point is parallel to the radius. This means that α is the same as the angle between the radius and the vertical direction.

In other words, α is the complement of the angle formed between the radius and the vertical direction. Let's denote that angle as β.

Since the bowling ball is a sphere, its radius is perpendicular to the surface at every point. Thus, β is the angle formed between the radius and the vertical direction.

Now, we have angle β and can calculate α using the expression α = 90° - β.

Finally, with the value of α determined, we can use the earlier expression tan(θ) = tan(α) to find the angle θ.

To summarize the process of finding the angle θ where the penny leaves the surface of the bowling ball:

1. Determine the angle β between the radius and the vertical direction.
2. Calculate α using α = 90° - β.
3. Solve for θ using the equation tan(θ) = tan(α).

Please note that in order to complete the calculations, you may need additional information such as the dimensions of the penny and the bowling ball.