Question

Beginning from rest when = 20°, a 35-kg child slides with negligible friction down the sliding board which is in the shape of a 2.5-m circular arc. Determine the tangential acceleration and speed of the child, and the normal force exerted on her (a) when = 30° and (b) when = 90°

Answer 1

Answer:

a) $a_{T}=8.50m/s^{2}$, $v_{T}=2.78 m/s$ and $N=279.83N$

b) $a_{T}=0m/s^{2}$, $v_{T}=5.68 m/s$ and $N=795.2N$

Explanation:

a)

In order to solve this problem we need to start by drawing a diagram of what the problem looks like: See attached picture. Next, we can start by finding the initial height of the child which will happen at an angle of 20°. We can find this by subtracting the distance from the highest point and the initial point from the radius of the circle so we get:

$h_{0}=2.5m-h_{1}$

so we get:

$h_{1}=2.5m(sin (20^{o}))$

$h_{1}=0.855m$

so

$h_{0}=2.5m-h_{1}$

$h_{0}=2.5m-0.855m$

$h_{0}=1.645m$

once we got this value, we can find the final height the same way. This time the angle is 30° so we get:

$h_{f}=2.5m-h_{2}$

$h_{f}=2.5m-2.5m(sin 30^{o}))$

$h_{f}=1.25m$

Once we have these heights, we can go ahead and use an energy balance equation to find the velocity at 30° so we get:

$U_{0}+K_{0}=U_{f}+K_{f}$

the initial kinetic energy is zero because its initial velocity is zero too, so the equation simplifies to:

$U_{0}=U_{f}+K_{f}$

so now we substitute with the corresponding formulas:

$mgh_{0}=mgh_{f}+\frac{1}{2}mv_{f}^{2}$

if we divided both sides of the equation by the mass, then the equation simplifies to:

$gh_{0}=gh_{f}+\frac{1}{2}v_{f}^{2}$

and now we can solve for the final velocity so we get:

$v_{f}=\sqrt{2g(h_{0}-h_{f}}$

and we can now substitute values:

$v_{f}=\sqrt{2(9.81m/s^{2})(1.645m-1.25m)}$

which solves to:

$v_{f}=2.78m/s$

which is our first answer. Once we got the velocity at 30° we can find the other data the problem is asking us for:

We can build a free body diagram (see attached  picture) and do a balance of forces so we get:

$\sum{F_{x}}=ma_{T}$

in this case we only have one x-force which is the x-component of the weight, so we get that:

$w_{x}=ma_{T}$

so we get:

$mgcos\theta=ma_{T}$

and solve for the tangential acceleration so we get:

$a_{T}=gcos\theta$

$a_{T}=9.81m/s^{2}*cos(30^{o})$

$a_{t}=8.50m/s^{2}$

Now we can find the centripetal acceleration by using the formula:

$a_{c}=\frac{V_{T}^{2}}{R}$

$a_{c}=\frac{(2.78m/s)^{2}}{2.5m}$

so we get:

$a_{c}=3.09m/s^{2}$

Next, we can find the normal force which is found by doing a sum of forces on y, so we get:

$\sum{F_{y}}=ma_{c}$

so we get:

$N-w_{y}=ma_{c}$

and we solve for the normal force so we get:

$N=ma_{c}+w_{y}$

and substitute:

$N=ma_{c}+mg sin\theta$

when factoring we get:

$N=m(a_{c}+g sin\theta)$

and we substitute:

$N=(35kg)(3.09m/s^{2}+9.81m/s^{2} sin30^{o})$

which yields:

N=279.83N

b)

The procedure for part b is mostly the same with some differences due to the angle. First:

$h_{0}=1.645m$

$h_{f}=0m$

so

$U_{0}+K_{0}=U_{f}+K_{f}$

in this case the initial kinetic energy is zero because the initial velocity is zero and the final potential energy is zero because the final height is zero as well, so the equation simplifies to:

$U_{0}=K_{f}$

so we get:

$mgh_{0}=\frac{1}{2}mv_{f}^{2}$

so we solve for the final velocity so we get:

$v_{f}=\sqrt{2gh_{0}}$

and we substitute:

$v_{f}=\sqrt{2(9.81m/s^{2})(1.645m)}$

$v_{f}=5.68m/s$

according to the free body diagram we get that:

$a_{T}=gcos\theta$

$a_{T}=9.81m/s^{2}(cos 90^{o})$

which yields:

$a_{T}=0$

we can also find the centripetal acceleration, so we get:

$a_{c}=\frac{V_{T}^{2}}{R}$

$a_{c}=\frac{(5.68m/s)^{2}}{2.5m}$

so we get:

$a_{c}=12.91m/s^{2}$

and we can do a sum of forces on y to find the normal force:

$\sum{F_{y}}=ma_{c}$

so we get:

$N-w_{y}=ma_{c}$

and we solve for the normal force so we get:

$N=ma_{c}+w_{y}$

and substitute:

$N=ma_{c}+mg$

when factoring we get:

$N=m(a_{c}+g)$

and we substitute:

$N=(35kg)(12.91m/s^{2}+9.81m/s^{2} sin30^{o})$

which yields:

N=795.2N

Answer 2

Following are the calculation to the angles:

For angle 30°:

consider the forces in tangential in direction:  

$\Sigma F_t= ma_t\\\\ W \cos \theta= m a_t \\\\m g \cos \theta = m a_t\\\\ a_t = g \cos \theta = 9.81 \cos 30^{\circ} \\\\a_t = 8.496 \ \frac{m}{s^2}$

calculate the speed of the child:

$g\cos \theta = a_t\\\\ g\cos \theta = \frac{v.dv}{R d\theta} \\\\g R \cos \theta d \theta= V \ dv$  

Integrating the equation:

$\int_{20}^{30} g R \cos \theta d \theta= \int_{0}^{v} V \ dv\\\\g R (\sin \theta)_{20}^{30} = [\frac{V^2}{2}]_{0}^{v} \ dv\\\\g R (\sin 30-\sin 20) = \frac{V^2}{2}\\\\9.81 \times 2.5 (\sin 30-\sin 20) = \frac{V^2}{2}\\\\v=2.79 \ \frac{m}{s}$

consider the forces in the normal declaration:  

$\Sigma F_n = ma_n \\\\N-mg \sin \theta= \frac{mv^2}{R} \\\\N- (35\times 9.81\times \sin 30) = \frac{35 \times 2.78^2}{2.5}\\\\N= 279.87\ N$

For angle 90°:

consider the forces in tangential in direction:  

$\Sigma F_t= ma_t\\\\ W \cos \theta= m a_t \\\\m g \cos \theta = m a_t\\\\ a_t = g \cos \theta = 9.81 \cos 90^{\circ} \\\\a_t = 0\ \frac{m}{s^2}$

calculate the speed of the child:  

$g\cos \theta = a_t\\\\ g\cos \theta = \frac{v.dv}{R d\theta} \\\\g R \cos \theta d \theta= V \ dv$  

Integrating the equation:

$\int_{20}^{30} g R \cos \theta d \theta= \int_{0}^{v} V \ dv\\\\g R (\sin \theta)_{20}^{90} = [\frac{V^2}{2}]_{0}^{v} \ dv\\\\g R (\sin 90-\sin 20) = \frac{V^2}{2}\\\\9.81 \times 2.5 (\sin 90-\sin 20) = \frac{V^2}{2}\\\\V=\sqrt{\frac{9.81 \times 2.5 (\sin 90-\sin 20) }{2}}$

   $=\sqrt{\frac{9.81 \times 2.5 (1 -0.342)}{2}}\\\\=\sqrt{8.06}\\\\=2.83$

consider the forces in the normal declaration:  

$\Sigma F_n = ma_n \\\\N-mg \sin \theta= \frac{mv^2}{R} \\\\N- (35\times 9.81\times \sin 90) = \frac{35 \times 2.78^2}{2.5}\\\\N= \frac{35 \times 2.78^2}{2.5 \times 35\times 9.81\times 1}$

    $= \frac{270.494}{858.375}\\\\=0.315$

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