Question

A car moving at 18.4 m/s encounters a bump that has a circular cross-section with a radius of 55.7 m. What is the normal force exerted by the seat of the car on a 63.0 kg passenger when the car is at the top of the bump?
A car moving at 18.4 m/s encounters a bump that has a circular cross-section with a radius of 55.7 m. What is the normal force exerted by the seat of the car on a 63.0 kg passenger when the car is at the top of the bump?
@Physics

Answer 1

Before I start answering this, is this a calculus based physics class or a non-calculus based physics class?

Answer 2

is a non-calculus based physics

Answer 3

I'm gonna assume calculus based, and we'll see what happens. If you think about an object moving up an incline at an angle theta with velocity v, the y-component of the velocity is equal to
\[v_y = v\sin(\theta)\]
A circular bump is really just like an incline of constantly changing angle, right? Anyways, if I want to know the acceleration in the y-direction, I just take the derivative of this object with respect to time. So,
\[a_y = \frac{dv_y}{dt} = v\cos(\theta)\frac{d\theta}{dt}\]
because of the chain rule. Furthermore, for an object moving along a circle, the rate of change of the angle is given by
\[\frac{d\theta}{dt} = \frac{v}{R}\]
So at any point along the bump, the net acceleration is
\[a_y = \frac{v^2\cos(theta)}{R} \]
making the net force
\[F = ma = \frac{mv^2\cos(\theta)}{R} \rightarrow F = \frac{mv^2}{R} \]
at the top. Summing the forces in the y-direction, we have that
\[F_{normal} - mg = \frac{mv^2}{R}\]
so finally we have
\[F_{normal} = \frac{mv^2}{R} + mg\]

Answer 4

Ah. Well, non-calculus based, you just have to start from
\[a = \frac{mv^2}{R} \]
for an object in circular motion. And this is an object in circular motion -- just a weird type of circular motion.

Answer 5

oops.
\[a = \frac{v^2}{R}\]

Answer 6

I just came across this again and I realize I made a boo-boo with the signs... it should be
\[mg - F_{normal} = \frac{mv^2}{r}\]
so
\[F_{normal} = mg-\frac{mv^2}{r} \]
sorry...

Answer 7

A stone is tied to a string (length = 1.05 m) and whirled in a circle at the same constant speed in two different ways. First, the circle is horizontal and the string is nearly parallel to the ground. Next, the circle is vertical. In the vertical case the maximum tension in the string is 14.6 % larger than the tension that exists when the circle is horizontal. Determine the speed of the stone.

Answer 8

pls help me with this as well

Answer 9

To keep an object moving in a circle of radius r at velocity v, it must be subjected to a force that is numerically equal to
\[F_c = \frac{mv^2}{r} \]
This may be provided by a number of different external forces. When your string is horizontal, then the tension in the string provides this force, so
\[T_H = \frac{mv^2}{r} \]
However, when the circle is vertical, there is an additional gravitational force acting downward that must be taken into account. It would not necessarily be easy to write down an equation describing the tension at some arbitrary angle, but you can convince yourself via common sense or physical intuition that the maximum tension will be when the string is at the bottom of the loop. At that point, summing the forces, we have that
\[T - mg = F_c = \frac{mv^2}{r} \]
and so
\[T_V = \frac{mv^2}{r} + mg = T_H + mg\]
Now, we're given that the maximum tension when the loop is vertical is 14.6% larger than when it's horizontal, so dividing by the horizontal tension:
\[\frac{T_V}{T_H} = 1+\frac{mg}{T_H} = 1.146 \rightarrow \frac{mg}{T_H} = 0.146\]
so we have that
\[T_H = \frac{mv^2}{r} = \frac{mg}{0.146} \rightarrow v = \sqrt{\frac{gr}{0.146}} = \sqrt{\frac{(9.8 m/s^2)(1.05m)}{0.146}} \approx 8.4\space m/s \]

Answer 10

thanks buddy,u must be a professor lol