Circular Motion

Circular motion requires a net (centripetal) force directed toward the centre, which continually changes the direction of the velocity. There is no real outward "centrifugal" force acting on the object.

The velocity is always tangent to the path, at right angles to the radius. The centripetal acceleration (toward the centre) changes the direction of this velocity without changing its magnitude.

Acceleration is any change in velocity, and velocity includes direction. Going around the roundabout the car's direction changes continuously, so it is accelerating (centripetally) even though its speed is constant.

$a = \dfrac{v^2}{r} = \dfrac{15^2}{45} = \dfrac{225}{45} = 5.0\text{ m/s}^2$, directed toward the centre of the bend.

In one period the ball travels one circumference, so $T = \dfrac{2\pi r}{v} = \dfrac{2\pi (0.60)}{3.0} \approx 1.3\text{ s}$.

First the angular velocity: $120\text{ rpm} = 2.0\text{ rev/s}$, so $\omega = 2.0 \times 2\pi \approx 12.6\text{ rad/s}$. Then $v = \omega r = 12.6 \times 0.30 \approx 3.8\text{ m/s}$.

$a = \dfrac{4\pi^2 r}{T^2} = \dfrac{4\pi^2 (0.25)}{(0.10)^2} = \dfrac{9.87}{0.01} \approx 990\text{ m/s}^2$ (about $100g$).

Balance means equal and opposite torques about the pivot: $m_1 g d_1 = m_2 g d_2$. The $g$ cancels: $30 \times 1.5 = 45 \times d$, so $d = \dfrac{45}{45} = 1.0\text{ m}$. The heavier child sits closer.

Torque uses the perpendicular component of the force: $\tau = rF\sin\theta = 0.50 \times 20 \times \sin 60^\circ = 10 \times 0.866 \approx 8.7\text{ N·m}$. (Maximum torque, $10\text{ N·m}$, occurs at $90^\circ$.)

The string tension provides the centripetal force, so at the breaking point $T = \dfrac{mv^2}{r}$: $v = \sqrt{\dfrac{Tr}{m}} = \sqrt{\dfrac{50 \times 1.0}{0.50}} = \sqrt{100} = 10\text{ m/s}$.

Friction supplies the centripetal force: $\mu mg = \dfrac{mv^2}{r}$, so $v = \sqrt{\mu rg} = \sqrt{0.50 \times 60 \times 9.8} = \sqrt{294} \approx 17\text{ m/s}$. The mass cancels.

Friction provides $m\omega^2 r$, maximum at $\mu mg$: $\mu mg = m\omega^2 r$, so $\omega = \sqrt{\dfrac{\mu g}{r}} = \sqrt{\dfrac{0.30 \times 9.8}{0.10}} = \sqrt{29.4} \approx 5.4\text{ rad/s}$.

The vertical component of tension supports the weight: $T\cos\theta = mg$, so $T = \dfrac{mg}{\cos\theta} = \dfrac{0.20 \times 9.8}{\cos 30^\circ} = \dfrac{1.96}{0.866} \approx 2.3\text{ N}$. The tension exceeds the weight because it must also supply the centripetal force.

The radius is $r = L\sin\theta = 0.80 \times 0.5 = 0.40\text{ m}$. Combining the force equations gives $v = \sqrt{rg\tan\theta} = \sqrt{0.40 \times 9.8 \times \tan 30^\circ} = \sqrt{2.26} \approx 1.5\text{ m/s}$.

On a frictionless banked curve the horizontal part of the normal force is the centripetal force, giving $\tan\theta = \dfrac{v^2}{rg} = \dfrac{20^2}{80 \times 9.8} = 0.510$, so $\theta \approx 27^\circ$.

Above the design speed, more centripetal force is needed than the normal force alone provides, so the car tends to slide up and out. Friction opposes this, acting down the slope toward the centre to supply the extra centripetal force.

At the minimum speed the tension is zero and gravity alone provides the centripetal force: $mg = \dfrac{mv^2}{r}$, so $v = \sqrt{gr} = \sqrt{9.8 \times 2.5} = \sqrt{24.5} \approx 4.9\text{ m/s}$.

It leaves the road when the normal force reaches zero and gravity alone supplies the centripetal force: $mg = \dfrac{mv^2}{r}$, so $v = \sqrt{gr} = \sqrt{9.8 \times 40} = \sqrt{392} \approx 20\text{ m/s}$.

The wall's normal force provides the centripetal force ($N = \frac{mv^2}{r}$), and friction $\mu N$ must support the weight: $\mu\dfrac{mv^2}{r} = mg$. So $v = \sqrt{\dfrac{rg}{\mu}} = \sqrt{\dfrac{3.0 \times 9.8}{0.40}} = \sqrt{73.5} \approx 8.6\text{ m/s}$.

Take torques about the right support. The beam's weight acts at its centre ($2.0\text{ m}$ from each end) and the load is $3.0\text{ m}$ from the right: $L \times 4.0 = 200 \times 2.0 + 300 \times 3.0 = 1300$, so $L = 325\text{ N}$.

The period is $T = 2\pi\sqrt{\dfrac{L\cos\theta}{g}}$. Rearranging, $\cos\theta = \dfrac{gT^2}{4\pi^2 L} = \dfrac{9.8 \times 4.0}{4\pi^2 \times 1.2} = \dfrac{39.2}{47.4} = 0.827$, so $\theta = \cos^{-1}(0.827) \approx 34^\circ$.

Energy conservation to the top (a rise of $2r$): $v_{top}^2 = v_{bot}^2 - 4gr = 64 - 4(9.8)(1.0) = 24.8$. At the top, $T + mg = \dfrac{mv_{top}^2}{r}$, so $T = m\left(\dfrac{v_{top}^2}{r} - g\right) = 0.50(24.8 - 9.8) = 7.5\text{ N}$.

At the top the minimum speed gives $v^2 = gr$. Energy conservation from height $h$ to the top (height $2r$): $gh = g(2r) + \tfrac{1}{2}v^2 = 2gr + \tfrac{1}{2}gr$, so $h = 2.5r = 2.5 \times 0.50 = 1.25\text{ m}$.

At the bottom $T_b = mg + \dfrac{mv_b^2}{r}$; at the top $T_t = \dfrac{mv_t^2}{r} - mg$. Energy conservation gives $v_b^2 - v_t^2 = 4gr$. Subtracting: $T_b - T_t = 2mg + \dfrac{m(v_b^2 - v_t^2)}{r} = 2mg + 4mg = 6mg$, independent of the speed.

At the top $v_t = \sqrt{gr}$, so $v_t^2 = gr$. Energy conservation over the drop of $2r$: $v_b^2 = v_t^2 + 2g(2r) = gr + 4gr = 5gr = 5(9.8)(1.0) = 49$, so $v_b = 7.0\text{ m/s}$.