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Initially, just before release, the system has potential energy $U$:

$$T=U=mgL$$

As the string hits the horizontal bar, it enters into a different 'orbit', this one with radius $L-H$. At the highest point of this loop, the mass is at height $2(L-H)$. Assuming full conservation of energy, then the new total energy is:

$$T=2mg(H-L)+K$$

Where $K$ is the kinetic energy at the top of the loop. So:

$$mgL=2mg(H-L)+K$$ So that:

$$K=mg(3L-2H)$$

With $v$ the tangential velocity at the top of the loop:

$$K=\frac12 mv^2=mg(3L-2H)$$

So:

$$v^2=2g(3L-2H)$$

The centripetal force $F_c$ is:

$$F_c=\frac{mv^2}{R}=\frac{mv^2}{L-H}$$

For there not to tension in the string at the top of the loop:

$$mg=\frac{mv^2}{L-H}$$

$$g=\frac{2g(3L-2H)}{L-H}$$

So I get:

$$L-H=2(3L-2H)$$

$$H=\frac53 L$$

Initially, just before release, the system has potential energy $U$:

$$T=U=mgL$$

As the string hits the horizontal bar, it enters into a different 'orbit', this one with radius $L-H$. At the highest point of this loop, the mass is at height $2(L-H)$. Assuming full conservation of energy, then the new total energy is:

$$T=2mg(L-H)+K$$

Where $K$ is the kinetic energy at the top of the loop. So:

$$mgL=2mg(L-H)+K$$ So that:

$$K=mg(2H-L)$$

With $v$ the tangential velocity at the top of the loop:

$$K=\frac12 mv^2=mg(2H-L)$$

So:

$$v^2=2g(2H-L)$$

The centripetal force $F_c$ is:

$$F_c=\frac{mv^2}{R}=\frac{mv^2}{L-H}$$

For there not to tension in the string at the top of the loop:

$$mg=\frac{mv^2}{L-H}$$

$$g=\frac{2g(2H-L)}{L-H}$$

So I get:

$$L-H=2(2H-L)$$

$$H=\frac35 L$$

Initially, just before release, the system has potential energy $U$:

$$T=U=mgL$$

As the string hits the horizontal bar, it enters into a different 'orbit', this one with radius $L-H$. At the highest point of this loop, the mass is at height $L-2(L-H)$. Assuming full conservation of energy, then the new total energy is:

$$T=2mg(H-L)+K$$

Where $K$ is the kinetic energy at the top of the loop. So:

$$mgL=2mg(H-L)+K$$ So that:

$$K=mg(3L-2H)$$

With $v$ the tangential velocity at the top of the loop:

$$K=\frac12 mv^2=mg(3L-2H)$$

So:

$$v^2=2g(3L-2H)$$

The centripetal force $F_c$ is:

$$F_c=\frac{mv^2}{R}=\frac{mv^2}{L-H}$$

For there not to tension in the string at the top of the loop:

$$mg=\frac{mv^2}{L-H}$$

$$g=\frac{2g(3L-2H)}{L-H}$$

So I get:

$$L-H=2(3L-2H)$$

$$H=\frac53 L$$

Initially, just before release, the system has potential energy $U$:

$$T=U=mgL$$

As the string hits the horizontal bar, it enters into a different 'orbit', this one with radius $L-H$. At the highest point of this loop, the mass is at height $2(L-H)$. Assuming full conservation of energy, then the new total energy is:

$$T=2mg(H-L)+K$$

Where $K$ is the kinetic energy at the top of the loop. So:

$$mgL=2mg(H-L)+K$$ So that:

$$K=mg(3L-2H)$$

With $v$ the tangential velocity at the top of the loop:

$$K=\frac12 mv^2=mg(3L-2H)$$

So:

$$v^2=2g(3L-2H)$$

The centripetal force $F_c$ is:

$$F_c=\frac{mv^2}{R}=\frac{mv^2}{L-H}$$

For there not to tension in the string at the top of the loop:

$$mg=\frac{mv^2}{L-H}$$

$$g=\frac{2g(3L-2H)}{L-H}$$

So I get:

$$L-H=2(3L-2H)$$

$$H=\frac53 L$$

Initially, just before release, the system has potential energy $U$:

$$T=U=mgL$$

As the string hits the horizontal bar, it enters into a different 'orbit', this one with radius $L-H$. At the highest point of this loop, the mass is at height $L-2(L-H)$. Assuming full conservation of energy, then the new total energy is:

$$T=2mg(L-H)+K$$

Where $K$ is the kinetic energy at the top of the loop. So:

$$mgL=2mg(L-H)+K$$ So that:

$$K=mg(2H-L)$$

With $v$ the tangential velocity at the top of the loop:

$$K=\frac12 mv^2=mg(2H-L)$$

So:

$$v^2=2g(2H-L)$$

The centripetal force $F_c$ is:

$$F_c=\frac{mv^2}{R}=\frac{mv^2}{L-H}$$

For there not to tension in the string at the top of the loop:

$$mg=\frac{mv^2}{L-H}$$

$$g=\frac{g(2H-L)}{L-H}$$

So I get:

$$L-H=2H-L$$

$$H=\frac23 H$$

The kinetic energy at the top of the loop is too high for the tension in the string to ever become zero.

Initially, just before release, the system has potential energy $U$:

$$T=U=mgL$$

As the string hits the horizontal bar, it enters into a different 'orbit', this one with radius $L-H$. At the highest point of this loop, the mass is at height $L-2(L-H)$. Assuming full conservation of energy, then the new total energy is:

$$T=2mg(H-L)+K$$

Where $K$ is the kinetic energy at the top of the loop. So:

$$mgL=2mg(H-L)+K$$ So that:

$$K=mg(3L-2H)$$

With $v$ the tangential velocity at the top of the loop:

$$K=\frac12 mv^2=mg(3L-2H)$$

So:

$$v^2=2g(3L-2H)$$

The centripetal force $F_c$ is:

$$F_c=\frac{mv^2}{R}=\frac{mv^2}{L-H}$$

For there not to tension in the string at the top of the loop:

$$mg=\frac{mv^2}{L-H}$$

$$g=\frac{2g(3L-2H)}{L-H}$$

So I get:

$$L-H=2(3L-2H)$$

$$H=\frac53 L$$