# Why is a bigger hammer more effective at hitting out bearings?

I was removing bearings from a motorcycle wheel. The bearings are fitted with a compression fit. I was trying to remove them with a regular hammer (~500g head)but the bearings wouldn't move. Using a lump hammer (3 kg head) was much more effective. In theory this should be Force = mass * acceleration. I was hitting the bearing as hard as I could with regular hammer and it wasn't moving, but just tapping with the lump hammer was more effective. So I feel there are other effects happening here. What could those be?

Assuming you apply the same work to both (Force * distance) then they will both gain equal kinetic energy. $$m_bV_b^2=m_sV_s^2$$ taking $$m_b = 6 m_s$$ so $$\sqrt{6}V_b = V_s$$ now let's calculate the ratio of momentum. $$\frac{m_bV_b}{m_sV_s} = \frac{6}{\sqrt{6}} \approx 2.45$$ What this means is that assuming they both have the same impact time, the larger hammer would produce 2.45 the average force of the smaller hammer.

The larger hammer has more inertia (resistance to acceleration) so transfers its momentum to whatever you hit. The lighter (less massive) hammer decelerates quickly when it meets the object youâ€™re hitting.

It can seem counter-intuitive, especially since a fast blow with the light hammer can have more KE than a slower blow with the heavier hammer yet still have less effect.

The reason is that less KE is transferred when a lighter hammer is used. In an elastic collision between two bodies, both momentum and KE must be conserved. When one body is very much lighter than the other, the two conditions can only be satisfied if the lighter body retains most of the KE through recoiling at virtually the same speed at which it approached the collision. If the difference in mass between the two colliding bodies is reduced, more KE is transferred from the lighter to the heavier, the transfer being a maximum when the masses of the bodies are equal. You can see this in a Newton's cradle, where the incoming sphere stops dead and the outgoing one carries off all the KE.

So, when you use a heavier hammer, it recoils less and more of its KE is transferred to the bearing/wheel. The frictional forces between the bearing and the wheel are insufficient to distribute the additional KE, so the bearing accelerates faster than the wheel and pops out.

It's about impulse. When you use a hammer with a head 6 times greater, the produced impulse exceeds 6 times for when you use a hammer with a lighter head. Forces of bodies interaction are very large and act for a short time. When the bodies contact each other, the force rises steeply and falls to zero again when they separate. the time of contact is very small $$\Delta t=t_f-t_i$$, say, around $$0.001\space s$$. The impulse $$\boldsymbol J$$ is the total change in momentum during the time interval $$t_f-t_i$$:

$$\boldsymbol J=\int\limits_{t_i}^{t_f}\boldsymbol F dt \space .$$

The time-average force over the mentioned time interval is defined by:

$$\boldsymbol F_{av}=\frac{1}{\Delta t}\int\limits_{t_i}^{t_f}\boldsymbol F dt=\frac{\boldsymbol J}{\Delta t} \space .$$

$$\boldsymbol F_{av}$$ is the constant force which gives the same impulse during $$\Delta t=t_f-t_i$$. In you problem, if we assume that $$\Delta t$$ is the same for both cases, and if the mass of the bearings and their velocity after contact are respectively $$m_b$$ and $$v_l$$ for when you use the lighter hammer, and $$m_b$$ and $$v_h$$ for when you use the heavier hammer, we can deduce:

$$\frac{\boldsymbol {(F_{av})}_h}{\boldsymbol {(F_{av})}_l}= \frac{\boldsymbol J_h/\Delta t}{\boldsymbol J_l/\Delta t}\approx\frac{m_bv_h}{m_bv_l}=\frac{v_h}{v_l} \space.$$

Therefore, if you want to calculate how much greater the average force is for when you use a heavier hammer, you need to measure the ratio of the velocities of the bearings for the very moments as they leave the hammer in the two cases.

Moreover, remember that a compression fit has static friction, and you need to produce sufficient force that overcomes the force of static friction. As long as the hammer's force does not reach the static friction force, the bearings do not move. As the exerted force becomes greater than the static friction force, say, by using a massive hammer, the coefficient of the static friction suddenly reduces to that of kinetic friction $$(\mu_k)$$, and the bearing accelerates. Considering the effect of friction may make the impulse calculations complicated. For instance, $$\boldsymbol {F_{av}}$$ is better to be replaced by $$\boldsymbol {F_{av}}-\mu_kpA$$, where $$p$$ is the static (compression) pressure and $$A$$ is the area of the bearing.

• I imagine its an elastic collision, both hammers are made of the same material(drop forged steel), so is the time of contact equal? – R. Doolan Nov 13 '19 at 10:21
• @R. Doolan The times of contacts are nearly equal, it may depend on many things. – Mohammad Javanshiry Nov 13 '19 at 12:18