Question

a) What is the maximum torque exerted by a 60 kg person riding a bike, if the rider puts all
his weight on each pedal when climbing a hill? The pedals rotate in a circle of radius
17 cm.
b) A ball of mass 1.5 kg rolling to the right with a speed of 3.6m/s, collides head-on with
a spring with a spring constant of 2.0Nm^-2.Determine the maximum compression of the
spring and the speed of the ball when the compression of the spring is 0.10 m.

Answer 1

Answer on Question #39587, Physics, Mechanics | Kinematics | Dynamics

a) What is the maximum torque exerted by a $60\mathrm{kg}$ person riding a bike, if the rider puts all his weight on each pedal when climbing a hill? The pedals rotate in a circle of radius $17~\mathrm{cm}$ .

b) A ball of mass $1.5 \, \text{kg}$ rolling to the right with a speed of $3.6 \, \text{m/s}$ , collides head-on with a spring with a spring constant of $2.0 \, \text{Nm}^{\wedge} - 2$ . Determine the maximum compression of the spring and the speed of the ball when the compression of the spring is $0.10 \, \text{m}$ .

Solution:

a) Torque, moment or moment of force, is the tendency of a force to rotate an object about an axis. Just as a force is a push or a pull, a torque can be thought of as a twist to an object. Mathematically, torque is defined as the cross product of the lever-arm distance and force, which tends to produce rotation.

The formula for torque is:

$$\tau = | \vec {r} \times \vec {F} | = r F \sin \theta$$

The word "maximum" implies that the force is applied at a $90{}^{\circ}$ angle to the radius, so the factor $\sin \theta$ becomes 1, and really the torque is:

$$\tau = r F$$

The force is the weight of the cyclist:

$$F = m g$$ $$F = (6 0 \mathrm {k g}) \cdot (9. 8 \mathrm {m} / \mathrm {s} ^ {2}) = 5 8 8 \mathrm {N}$$ $$r = 0. 1 7 \mathrm {m}$$

Thus

$$\tau = r F = 0. 1 7 \cdot 5 8 8 = 9 9. 9 6 \mathrm {N} \cdot \mathrm {m}$$

b)

Conservation of energy tells us that the kinetic energy of the ball is equal to the potential energy of the spring

$$\begin{array}{c} \mathrm {K E} = \mathrm {P E} \\ \frac {m v ^ {2}}{2} = \frac {k x _ {\text {m a x}} ^ {2}}{2} \\ x _ {\text {m a x}} = v \sqrt {\frac {m}{k}} = 3. 6 \sqrt {\frac {1 . 5}{2 . 0}} = 3. 1 2 \mathrm {m}. \end{array}$$

The speed of the ball when the compression of the spring is $0.10 \, \text{m}$:

$$v_x = v - x \sqrt{\frac{k}{m}} = 3.6 - 0.1 \sqrt{\frac{2.0}{1.5}} = 3.484 \approx 3.48 \, \text{m/s}$$

Answer. a) $\tau = 99.96 \, \text{N} \cdot \text{m}$;

b) $x_{max} = 3.12 \, \text{m}, v_x = 3.48 \, \text{m/s}$.