A hammer thrower accelerates the hammer\(\left( {{\bf{mass}} = {\bf{7}}{\bf{.30 kg}}} \right)\)from rest within four full turns (revolutions) and releases it at a speed of\({\bf{26}}{\bf{.5 m/s}}\).Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius 1.20 m, calculate (a) the angular acceleration, (b) the (linear) tangential acceleration, (c) the centripetal acceleration just before release, (d) the net force being exerted on the hammer by the athlete just before release, and (e) the angle of this force with respect to the radius of the circular motion. Ignore gravity.

The hammer moved in a circular motion when a hammer thrower threw it. The acceleration of the hammer can be obtained by the resultant of the angular and translational accelerations.From Newton’s second law, the net force acting on the hammer can be evaluated.

The given data can be listed below as:

• The mass of the hammer is\(m = 7.30{\rm{ kg}}\).
• The initial angular speed ofthe hammeris\({\omega _i} = 0{\rm{ rad/s}}\).
• The revolution of the hammer is\(\theta = 4{\rm{ rev}}\left( {\frac{{2\pi {\rm{ rad}}}}{{1{\rm{ rev}}}}} \right) = 8\pi {\rm{ rad}}\).
• The radius of the sphere is \(r = 1.20{\rm{ m}}\).
• The initial speed of the hammer is\({v_i} = 0{\rm{ m/s}}\).
• The final speed of the hammer is\({v_f} = 26.5{\rm{ m/s}}\).
• The acceleration due to gravity is\(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

The accelerations can be represented as:

Here,\({a_n}\)is thenet acceleration,\({a_t}\)is the tangential acceleration,\({a_r}\)is the radial or centripetal acceleration,\(\theta \)is the angle between both accelerations, which is also equal to the angle of the force exerted on the hammer with respect to the radius of the circular motion.

The final angular speed of the hammer can be expressed as:

\({\omega _f} = \frac{{{v_f}}}{r}\)

Substitute the values in the above equation.

\(\begin{align}{\omega _f} &= \frac{{{\rm{26}}{\rm{.5 m/s}}}}{{{\rm{1}}{\rm{.20 m}}}}\\ &= 22.08{\rm{ rad/s}}\end{align}\)

The angular acceleration of the hammer can be expressed as:

\(\alpha = \frac{{\omega _f^2 - \omega _i^2}}{{2\theta }}\)

Substitute the values in the above equation.

\(\begin{align}\alpha &= \frac{{{{\left( {22.08{\rm{ rad/s}}} \right)}^2} - {{\left( {0{\rm{ rad/s}}} \right)}^2}}}{{2 \times 8\pi {\rm{ rad}}}}\\ &= 9.7{\rm{ rad/}}{{\rm{s}}^2}\end{align}\)

Thus, the angular acceleration of the hammer is \(9.7{\rm{ rad/}}{{\rm{s}}^2}\).

The tangential acceleration of the hammer can be expressed as:

\({a_t} = r\alpha \)

Substitute the values in the above equation.

\(\begin{align}{a_t} &= 1.20{\rm{ m}} \times 9.7{\rm{ rad/}}{{\rm{s}}^2}\\ &= 11.64{\rm{ m/}}{{\rm{s}}^2}\end{align}\)

Thus, the linear acceleration of the hammer is \(11.64{\rm{ m/}}{{\rm{s}}^2}\).

The centripetal or radial acceleration of the hammer can be expressed as:

\(\begin{align}{a_c} &= {a_r}\\ &= \omega _f^2r\\ &= \frac{{v_f^2}}{r}\end{align}\)

Substitute the values in the above equation.

\(\begin{align}{a_r} &= \frac{{{{\left( {{\rm{26}}{\rm{.5 m/s}}} \right)}^2}}}{{{\rm{1}}{\rm{.20 m}}}}\\ &= 585.21{\rm{ m/}}{{\rm{s}}^2}\end{align}\)

Thus, the centripetal acceleration of the hammer is \(585.21{\rm{ m/}}{{\rm{s}}^2}\).

The net acceleration of the hammer can be expressed as:

\({a_n} = \sqrt {a_t^2 + a_r^2} \)

Substitute the values in the above equation.

\(\begin{align}{a_n} &= \sqrt {{{\left( {11.64{\rm{ m/}}{{\rm{s}}^2}} \right)}^2} + {{\left( {585.21{\rm{ m/}}{{\rm{s}}^2}} \right)}^2}} \\ &= 585.33{\rm{ m/}}{{\rm{s}}^2}\end{align}\)

The net force exerted on the hammer can be expressed as:

\(F = m{a_n}\)

Substitute the values in the above equation.

\(\begin{align}F &= 7.30{\rm{ kg}} \times 585.33{\rm{ m/}}{{\rm{s}}^2}\\ &= 4272.91{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ N}}}}{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\ &= 4272.91{\rm{ N}}\end{align}\)

Thus, the net force exerted on the hammer is \(4272.91{\rm{ N}}\).

The angle of this force with respect to the radius can be expressed as:

\(\begin{align}\tan \theta &= \left( {\frac{{{a_t}}}{{{a_r}}}} \right)\\\theta &= {\tan ^{ - 1}}\left( {\frac{{{a_t}}}{{{a_r}}}} \right)\end{align}\)

Substitute the values in the above equation.

\(\begin{align}\theta &= {\tan ^{ - 1}}\left( {\frac{{11.64{\rm{ m/}}{{\rm{s}}^2}}}{{585.21{\rm{ m/}}{{\rm{s}}^2}}}} \right)\\ &= {\tan ^{ - 1}}\left( {0.0199} \right)\\ &= 1.14^\circ \end{align}\)

Thus, the angle of this force with respect to the radius of the circular motion is \(1.14^\circ \).